I'm currently writing a function that should basically just write characters from a string into variables.
When performing test prints my variables seem fine. But when I attempt to print the first variable assigned(inchar) outside of the function it returns a empty string, but the second variable (outchar) seems to return fine. Am I somehow overwriting the first variable?
This is my code:
_EXIT = 1
_READ = 3
_WRITE = 4
_STDOUT = 1
_STDIN = 1
_GETCHAR = 117
MAXBUFF = 100
.SECT .TEXT
start:
0: PUSH endpro2-prompt2
PUSH prompt2
PUSH _STDOUT
PUSH _WRITE
SYS
ADD SP,8
PUSH 4
PUSH buff
CALL getline
ADD SP,4
!!!!!!!!!
PUSH buff
CALL gettrans
ADD SP,4
ADD AX,1 !gives AX an intial value to start loop
1: CMP AX,0
JE 2f
PUSH endpro-prompt1
PUSH prompt1
PUSH _STDOUT
PUSH _WRITE
SYS
ADD SP,8
PUSH MAXBUFF
PUSH buff
CALL getline
ADD SP,2
!PUSH buff
!CALL translate
!ADD SP,4
JMP 1b
2: PUSH 0 ! exit with normal exit status
PUSH _EXIT
SYS
getline:
PUSH BX
PUSH CX
PUSH BP
MOV BP,SP
MOV BX,8(BP)
MOV CX,8(BP)
ADD CX,10(BP)
SUB CX,1
1: CMP CX,BX
JE 2f
PUSH _GETCHAR
SYS
ADD SP,2
CMPB AL,-1
JE 2f
MOVB (BX),AL
INC BX
CMPB AL,'\n'
JNE 1b
2: MOVB (BX),0
MOV AX, BX
SUB AX,8(BP)
POP BP
POP CX
POP BX
RET
gettrans:
PUSH BX
PUSH BP
MOV BP,SP
MOV BX,6(BP) !Store argument in BX
MOVB (inchar),BL ! move first char to inchar
1: INC BX
CMPB (BX),' '
JE 1b
MOVB (outchar),BL !Move char seperated by Space to outchar
MOV AX,1 !On success
POP BP
POP BX
RET
.SECT .BSS
buff:
.SPACE MAXBUFF
.SECT .DATA
prompt1:
.ASCII "Enter a line of text: "
endpro:
prompt2:
.ASCII "Enter 2 characters for translation: "
endpro2:
outchar:
.BYTE 0
inchar:
.BYTE 0
charct:
.BYTE 0
wordct:
.BYTE 0
linect:
.BYTE 0
inword:
.BYTE 0
This is the code used to test print
PUSH 1 ! print that byte
PUSH inchar
PUSH _STDOUT
PUSH _WRITE
SYS
ADD SP,8
CALL printnl !function that prints new line
PUSH 1 ! print that byte
PUSH outchar
PUSH _STDOUT
PUSH _WRITE
SYS
CALL printnl
ADD SP,8
There seem to be a number of as88 8088 simulator environments. But I noticed on many of the repositories of code this bug mentioned:
1. The assembler requires sections to be defined in the following order:
TEXT
DATA
BSS
After the first occurrences, remaining section directives may appear in any order.
I'd recommend in your code to move the BSS section after DATA in the event your as88 environment has a similar problem.
In your original code you had lines like this:
MOV (outchar),BX
[snip]
MOV (inchar),BX
You defined outchar and inchar as bytes. The 2 lines above move 2 bytes (16-bits) from the BX register to both one byte variables. This will cause the CPU to write the extra byte into the next variable in memory. You'd want to explicitly move a single byte. Something like this might have been more appropriate:
MOVB (outchar),BL
[snip]
MOVB (inchar),BL
As you will see this code still has a bug as I mention later in this answer. To clarify - the MOVB instruction will move a single byte from BL and place it into the variable.
When you do a SYS call for Write you need to pass the address of the buffer to print, not the data in the buffer. You had 2 lines like this:
PUSH (inchar)
[snip]
PUSH (outchar)
The parentheses say to take the value in the variables and push them on the stack. SYS WRITE requires the address of the characters to display. The code to push their addresses should look like:
PUSH inchar
[snip]
PUSH outchar
gettrans function has a serious flaw in handling the copy of a byte from one buffer to another. You have code that does this:
MOV BX,6(BP) !Store argument in BX
MOVB (inchar),BL ! move first char to inchar
1: INC BX
CMPB (BX),' '
JE 1b
MOVB (outchar),BL !Move char seperated by Space to outchar
MOV BX,6(BP) properly places that buffer address passed as an argument and puts it into BX. There appears to be a problem with the lines that look like:
MOVB (inchar),BL ! move first char to inchar
This isn't doing what the comment suggests it should. The line above moves the lower byte (BL) of the buffer address in BX to the variable inchar . You want to move the byte at the memory location pointed to by BX and put it into inchar. Unfortunately on the x86 you can't move the data from one memory operand to another directly. To get around this you will have to move the data from the buffer pointed to by BX into a temporary register (I'll choose CL) and then move that to the variable. The code could look like this:
MOVB CL, (BX)
MOVB (inchar),CL ! move first char to inchar
You then have to do the same for outchar so the fix in both places could look similar to this:
MOV BX,8(BP) !Store argument in BX
MOVB CL, (BX)
MOVB (inchar),CL ! move first char to inchar
1: INC BX
CMPB (BX),' '
JE 1b
MOVB CL, (BX)
MOVB (outchar),CL ! move second char to outchar
The instruction MOV (inchar),BX stores register BX to the memory location labelled inchar.
However, inchar has been defined as a .BYTE, but BX is a 16-bit register, (2 bytes,) so you are writing not only inchar but also outchar.
The only reason why it appears to work in the beginning is because the 8088 is a low-endian architecture, so the low-order byte of BX is being stored first, while the high-order byte follows.
So, try MOV (inchar),BL
I'm starting to read LLVM docs and IR documentation.
In common architectures, an asm cmp instruction "result" value is -at least- 3 bits long, let's say the first bit is the SIGN flag, the second bit is the CARRY flag and the third bit is the ZERO flag.
Question 1)
Why the IR icmp instruction result value is only i1? (you can choose only one flag)
Why doesn't IR define, let's call it a icmp2 instruction returning an i3 having SIGN,CARRY and ZERO flags?
This i3 value can be acted upon with a switch instruction, or maybe a specific br2 instruction, like:
%result = cmp2 i32 %a, i32 %b
br2 i3 %result onzero label %EQUAL, onsign label %A_LT_B
#here %a GT %b
Question 2)
Does this make sense? Could this br2 instruction help create new optimizations? i.e. remove all jmps? it is necessary or the performance gains are negligible?
The reason I'm asking this -besides not being an expert in LLVM- is because in my first tests I was expecting some kind of optimization to be made by LLVM in order to avoid making the comparison twice and also avoid all branches by using asm conditional-move instructions.
My Tests:
I've compiled with clang-LLVM this:
#include <stdlib.h>
#include <inttypes.h>
typedef int32_t i32;
i32 compare (i32 a, i32 b){
// return (a - b) & 1;
if (a>b) return 1;
if (a<b) return -1;
return 0;
}
int main(int argc, char** args){
i32 n,i;
i32 a,b,avg;
srand(0); //fixed seed
for (i=0;i<500;i++){
for (n=0;n<1e6;n++){
a=rand();
b=rand();
avg+=compare(a,b);
}
}
return avg;
}
Output asm is:
...
mov r15d, -1
...
.LBB1_2: # Parent Loop BB1_1 Depth=1
# => This Inner Loop Header: Depth=2
call rand
mov r12d, eax
call rand
mov ecx, 1
cmp r12d, eax
jg .LBB1_4
# BB#3: # in Loop: Header=BB1_2 Depth=2
mov ecx, 0
cmovl ecx, r15d
.LBB1_4: # %compare.exit
# in Loop: Header=BB1_2 Depth=2
add ebx, ecx
...
I expected (all jmps removed in the inner loop):
mov r15d, -1
mov r13d, 1 # HAND CODED
call rand
mov r12d, eax
call rand
xor ecx,ecx # HAND CODED
cmp r12d, eax
cmovl ecx, r15d # HAND CODED
cmovg ecx, r13d # HAND CODED
add ebx, ecx
Performance difference (1s) seems to be negligible (on a VM under VirtualBox):
LLVM generated asm: 12.53s
hancoded asm: 11.53s
diff: 1s, in 500 millions iterations
Question 3)
Are my performance measures correct? Here's the makefile and the full hancoded.compare.s
makefile:
CC=clang -mllvm --x86-asm-syntax=intel
all:
$(CC) -S -O3 compare.c
$(CC) compare.s -o compare.test
$(CC) handcoded.compare.s -o handcoded.compare.test
echo `time ./compare.test`
echo `time ./handcoded.compare.test`
echo `time ./compare.test`
echo `time ./handcoded.compare.test`
hand coded (fixed) asm:
.text
.file "handcoded.compare.c"
.globl compare
.align 16, 0x90
.type compare,#function
compare: # #compare
.cfi_startproc
# BB#0:
mov eax, 1
cmp edi, esi
jg .LBB0_2
# BB#1:
xor ecx, ecx
cmp edi, esi
mov eax, -1
cmovge eax, ecx
.LBB0_2:
ret
.Ltmp0:
.size compare, .Ltmp0-compare
.cfi_endproc
.globl main
.align 16, 0x90
.type main,#function
main: # #main
.cfi_startproc
# BB#0:
push rbp
.Ltmp1:
.cfi_def_cfa_offset 16
push r15
.Ltmp2:
.cfi_def_cfa_offset 24
push r14
.Ltmp3:
.cfi_def_cfa_offset 32
push r12
.Ltmp4:
.cfi_def_cfa_offset 40
push rbx
.Ltmp5:
.cfi_def_cfa_offset 48
.Ltmp6:
.cfi_offset rbx, -48
.Ltmp7:
.cfi_offset r12, -40
.Ltmp8:
.cfi_offset r14, -32
.Ltmp9:
.cfi_offset r15, -24
.Ltmp10:
.cfi_offset rbp, -16
xor r14d, r14d
xor edi, edi
call srand
mov r15d, -1
mov r13d, 1 # HAND CODED
# implicit-def: EBX
.align 16, 0x90
.LBB1_1: # %.preheader
# =>This Loop Header: Depth=1
# Child Loop BB1_2 Depth 2
mov ebp, 1000000
.align 16, 0x90
.LBB1_2: # Parent Loop BB1_1 Depth=1
# => This Inner Loop Header: Depth=2
call rand
mov r12d, eax
call rand
xor ecx,ecx #hand coded
cmp r12d, eax
cmovl ecx, r15d #hand coded
cmovg ecx, r13d #hand coded
add ebx, ecx
.LBB1_3:
dec ebp
jne .LBB1_2
# BB#5: # in Loop: Header=BB1_1 Depth=1
inc r14d
cmp r14d, 500
jne .LBB1_1
# BB#6:
mov eax, ebx
pop rbx
pop r12
pop r14
pop r15
pop rbp
ret
.Ltmp11:
.size main, .Ltmp11-main
.cfi_endproc
.ident "Debian clang version 3.5.0-1~exp1 (trunk) (based on LLVM 3.5.0)"
.section ".note.GNU-stack","",#progbits
Question 1: LLVM IR is machine independent. Some machines might not even have a carry flag, or even a zero flag or sign flag. The return value is i1 which suffices to indicate TRUE or FALSE. You can set the comparison condition like 'eq' and then check the result to see if the two operands are equal or not, etc.
Question 2: LLVM IR does not care about optimization initially. The main goal is to generate a Static Single Assignment (SSA) based representation of instructions. Optimization happens in later passes of which some are machine independent and some are machine dependent. Your br2 idea will assume that the machine will support those 3 flags which might be a wrong assumption,
Question 3: I am not sure what you are trying to do here. Can you explain more?
I wanted to print the first 20 numbers using loop.
Printing the first nine numbers is absolutely fine as the hexadecimal and decimal codes are the same, but from the 10th number I had to convert each number into its appropriate code and then convert it and store it to string and eventually display it
That is,
If (NUMBER > 9)
ADD 6D
;10d = 0ah --(+6)--> 16d = 10h
IF NUMBER IS > 19
ADD 12D
;20d = 14h --(+12)--> 32d = 20h
Then rotating and shifting each number to get the desired output number, that is,
DAA # let al = 74h = 0111.0100
XOR AH,AH # ah = 0 (Just in case it wasn't)
# ax = 0000.0000.0111.0100
ROR AX,4 # ax = 0100.0000.0000.0111 = 4007h
SHR AH,4 # ax = 0000.0100.0000.0111 = 0407h
ADD AX,3030h # ax = 0011.0100.0011.0111 = 3437h = ASCII "74" (Reversed due to little endian)
And then storing the result in to the string and displaying it, that is,
MOV BX,OFFSET Result ;Let Result is an empty string
MOV byte ptr[BX],5 ;Size of the string
MOV byte ptr[BX+4],'$' ;String terminator
MOV byte ptr[BX+3],AH ;storing number
MOV byte ptr[BX+2],AL
MOV DX,BX
ADD DX,02 ;Displaying the result
MOV AH,09H ;Interrupt 21 service to display string
INT 21H
And here is the complete code with proper commenting,
MOV CX,20 ;Number of iterations
MOV DX,0 ;First value of the sequence
L1:
PUSH DX
ADD DX,30H ; 30H is equal to 0 in hexadecimal , 31H = 1 and so on
MOV AH,02H ; INTERRUPT Service to print the DX content
INT 21H
POP DX
ADD DX,1
CMP DX,09 ; if number is > 9 i.e 0A then go to L2
JA L2
LOOP L1
L2:
PUSH DX
MOV AX,DX
CMP AX,14H ;If number is equal to 14H(20) then Jump to L3
JE L3
ADD AX,6D ;If less than 20 then add 6D
XOR AH,AH ;Clear the content of AH
ROR AX,4 ;Rotating and Shifting for to properly store
SHR AH,4
ADC AX,3030h
MOV BX,OFFSET Result
MOV byte ptr[BX],5
MOV byte ptr[BX+4],'$'
MOV byte ptr[BX+3],AH
MOV byte ptr[BX+2],AL
MOV DX,BX
ADD DX,02
MOV AH,09H
INT 21H
POP DX
ADD DX,1
LOOP L2
;If the number is equal to 20 come here, ->
; Every step is repeated here just to change 6D to 12D
L3:
ADD AX,12D
XOR AH,AH
ROR AX,1
ROR AX,1
ROR AX,1
ROR AX,1
SHR AH,1
SHR AH,1
SHR AH,1
SHR AH,1
ADC AX,3030h
MOV BX,OFFSET Result
MOV byte ptr[BX],5
MOV byte ptr[BX+4],'$'
MOV byte ptr[BX+3],AH
MOV byte ptr[BX+2],AL
MOV DX,BX
ADD DX,02
MOV AH,09H
INT 21H
Is there any proper way to do it, creating a function and using if/else (jumps) to get the desired output rather than repeating the code again and again?
PSEUDO CODE:
VAR = 6
IF Number is > 9
ADD AX,VAR
Else IF Number is > 19
ADD AX,(VAR*2)
ELSE IF NUMBER is > 29
ADD AX,(VAR*3)
So you just want to print 0 ... 20 as ASCII characters? It looks like you understand that the numerals are identified as 0x30 ... 0x39 for '0' to '9', so you could use integer division to generate the character for the tens digit:
I usually work with C but conversion to assembler shouldn't be too complicated since these are all fundamental operations and there are no function calls.
int i_value = 29;
int i_tens = i_value/10; //Integer division! 29/10 = 2, save for later use
char c_tens = '0' + i_tens;
char c_ones = '0' + i_value-(10*i_tens); // Subtract N*10 from value
The output will be c_tens = 0x32, c_ones = 0x39. You should be able to wrap this inside of a loop pretty easily using a pair of registers.
Pseudocode
regA <- num_iterations //For example, 20
regB <- 0 //Initialize counter register
LOOP:
//Do conversion for the current iteration.
//Manipulate bytes for output as necessary.
regB <- regB +1
branch not equal regA, regB LOOP
The following code counts from 0 up to 99 (ax contains the ASCII number):
count proc
mov cx, 100 ; loop runs the times specified in the cx register
xor bx, bx ; set counter to zero
print:
mov ax, bx
aam ; Converts binary to unpacked BCD
xor ax, 3030h ; Converts upacked BCD to ASCII
; Print here (ax now contains the numer in ASCII representation)
inc bx ; Increase counter
loop print
ret
count endp
I am implementing a uart queue in s3c44b0x (ARM7TDMI), the uart0 ISR will enqueue the char while the main loop will dequeue the char. however, while dequeuing, the value (in R0) returned may be not the one dequeued from the queue, and I found R0 is violated after returning from the dequeue function (input 'v' cont., and test() is in the main loop):
wish for your help.
CHAR cliDequeue(void)
{
CHAR bTmpCh;
if (gwCliQSize == 0)
{
return 0;
}
bTmpCh = gabCliQ[gwCliQTail]; /* char is enqueued in the Q in ISR */
gwCliQTail++;
gwCliQTail %= MAX_CLI_QUEUE_LEN;
ASSERT(gwCliQSize > 0);
gwCliQSize--;
ASSERT(bTmpCh == 'v'); /* will not assert */
//uartPutChar(bTmpCh);
return bTmpCh;
}
void test(void)
{
CHAR bTestCh;
bTestCh = cliDequeue();
if (bTestCh != 0)
{
ASSERT(bTestCh == 'v'); /* assert here ! */
uartPutChar(bTestCh);
}
}
We don't have enough information / context to answer definitively. It would also be helpful if you posted the corresponding assembly code so that we could see how/when things are moved in & out of R0. REgardless, a few things spring to mind immediately from your posted C code.
(0) Are the variables shared between interrupts & the main loop declared as volatile?
(1)
In CliDequeue, you're accessing an array which is shared with an ISR. It appears to be a single reader / single writer construct, so that isn't automatically bad, but your housekeeping isn't airtight.
For example, one invariant you must be sure to satisfy is that the queue size & tail pointer are in sync. Yet, unless this routine is called with interrupts disabled, your tail pointer & queue size aren't adjusted as a single transaction.
(2)
Furthermore, I'd guess that gwCliQSize is also adjusted in the interrupt (incremented in ISR, decremented in the application). Another race condition. To perform gwCliQSize--, behind the scenes you are probably reading from memory to a register, decrementing the register, then writing it back. What happens if you read 5 from memory into R1, then an interrupt fires and increments it to 6, then you exit the ISR, and the register decrement and writeback (with a value of 4).
(3)
Lastly, it's possible (although not too likely) that bTmpCh or bTestCh are stored on the stack, and that your stack is getting corrupted / slammed by another task / interrupt / etc. So when your assert fails, you're thinking it's R0 that is corrupted, but really it could be that the value moved into R0 before return, or the value moved out of R0 into a stack variable, is getting clobbered.
I've nattered on enough. There are other possibilities but from what you've posted (and not posted) it's impossible to say for sure.
P.S. If you've used a debugger and it's really & literally R0's value that is getting corrupted, not just the value of the character in the queue, that points to a problem in your scheduler / context switcher / ISR pre- or post-amble etc...
here is the assembly code:
for the test():
0x00001308 E92D4010 STMDB R13!,{R4,R14}
37: bTestCh = cliDequeue();
38:
0x0000130C EB000207 BL cliDequeue(0x00001B30)
0x00001310 E1A04000 MOV R4,R0
39: if (bTestCh != 0)
40: {
0x00001314 E3540000 CMP R4,#pTest(0x00000000)
0x00001318 0A000007 BEQ 0x0000133C
41: ASSERT(bTestCh == 'v');
0x0000131C E1A00000 NOP
0x00001320 E3540076 CMP R4,#0x00000076
0x00001324 0A000001 BEQ 0x00001330
0x00001328 E1A00000 NOP
0x0000132C EAFFFFFE B 0x0000132C
0x00001330 E1A00000 NOP
42: uartPutChar(bTestCh);
43: }
0x00001334 E1A00004 MOV R0,R4
0x00001338 EB00014A BL uartPutChar(0x00001868)
44: }
45:
46: int main(void)
0x0000133C E8BD4010 LDMIA R13!,{R4,R14}
0x00001340 E12FFF1E BX R14
for the cliDequeu(), BTW, gwCliQSize is defined as
UINT32 volatile gwCliQSize;
0x00001B30 E59F00D4 LDR R0,[PC,#0x00D4]
0x00001B34 E5900000 LDR R0,[R0]
0x00001B38 E3500000 CMP R0,#pTest(0x00000000)
0x00001B3C 1A000001 BNE 0x00001B48
78: return 0;
79: }
80:
81: bTmpCh = gabCliQ[gwCliQTail];
82: gwCliQTail++;
83: gwCliQTail %= MAX_CLI_QUEUE_LEN;
84: ASSERT(gwCliQSize > 0);
85: gwCliQSize--;
86:
87: //chCheck(bTmpCh);
88: ASSERT(bTmpCh == 'v'); /* will not assert */
89: //uartPutChar(bTmpCh);
90:
91: return bTmpCh;
0x00001B40 E3A00000 MOV R0,#pTest(0x00000000)
92: }
93:
94:
95: void cliQInit(void)
0x00001B44 E12FFF1E BX R14
81: bTmpCh = gabCliQ[gwCliQTail];
0x00001B48 E59F00C0 LDR R0,[PC,#0x00C0]
0x00001B4C E59F20C4 LDR R2,[PC,#0x00C4]
0x00001B50 E5922000 LDR R2,[R2]
0x00001B54 E7D01002 LDRB R1,[R0,R2]
82: gwCliQTail++;
0x00001B58 E59F00B8 LDR R0,[PC,#0x00B8]
0x00001B5C E5900000 LDR R0,[R0]
0x00001B60 E2800001 ADD R0,R0,#0x00000001
0x00001B64 E59F20AC LDR R2,[PC,#0x00AC]
0x00001B68 E5820000 STR R0,[R2]
83: gwCliQTail %= MAX_CLI_QUEUE_LEN;
0x00001B6C E2820000 ADD R0,R2,#pTest(0x00000000)
0x00001B70 E5900000 LDR R0,[R0]
0x00001B74 E20000FF AND R0,R0,#0x000000FF
0x00001B78 E5820000 STR R0,[R2]
84: ASSERT(gwCliQSize > 0);
0x00001B7C E1A00000 NOP
0x00001B80 E59F0084 LDR R0,[PC,#0x0084]
0x00001B84 E5900000 LDR R0,[R0]
0x00001B88 E3500000 CMP R0,#pTest(0x00000000)
0x00001B8C 1A000001 BNE 0x00001B98
0x00001B90 E1A00000 NOP
0x00001B94 EAFFFFFE B 0x00001B94
0x00001B98 E1A00000 NOP
85: gwCliQSize--;
86:
87: //chCheck(bTmpCh);
0x00001B9C E59F0068 LDR R0,[PC,#0x0068]
0x00001BA0 E5900000 LDR R0,[R0]
0x00001BA4 E2400001 SUB R0,R0,#0x00000001
0x00001BA8 E59F205C LDR R2,[PC,#0x005C]
0x00001BAC E5820000 STR R0,[R2]
88: ASSERT(bTmpCh == 'v'); /* will not assert */
89: //uartPutChar(bTmpCh);
90:
0x00001BB0 E1A00000 NOP
0x00001BB4 E3510076 CMP R1,#0x00000076
0x00001BB8 0A000001 BEQ 0x00001BC4
0x00001BBC E1A00000 NOP
0x00001BC0 EAFFFFFE B 0x00001BC0
0x00001BC4 E1A00000 NOP
91: return bTmpCh;
92: }
93:
94:
95: void cliQInit(void)
0x00001BC8 E1A00001 MOV R0,R1
0x00001BCC EAFFFFDC B 0x00001B44
for cliEnqueue:
void cliEnqueue(CHAR bC)
{
if (gwCliQSize == MAX_CLI_QUEUE_LEN)
{
ASSERT(0);
}
gabCliQ[gwCliQHeader] = bC;
gwCliQHeader++;
gwCliQHeader %= MAX_CLI_QUEUE_LEN;
gwCliQSize++;
}
assembly:
0x00001A5C E59F11AC LDR R1,[PC,#0x01AC]
0x00001A60 E59F21AC LDR R2,[PC,#0x01AC]
0x00001A64 E5922000 LDR R2,[R2]
0x00001A68 E7C10002 STRB R0,[R1,R2]
25: gwCliQHeader++;
0x00001A6C E59F11A0 LDR R1,[PC,#0x01A0]
0x00001A70 E5911000 LDR R1,[R1]
0x00001A74 E2811001 ADD R1,R1,#0x00000001
0x00001A78 E59F2194 LDR R2,[PC,#0x0194]
0x00001A7C E5821000 STR R1,[R2]
26: gwCliQHeader %= MAX_CLI_QUEUE_LEN;
0x00001A80 E2821000 ADD R1,R2,#pTest(0x00000000)
0x00001A84 E5911000 LDR R1,[R1]
0x00001A88 E20110FF AND R1,R1,#0x000000FF
0x00001A8C E5821000 STR R1,[R2]
27: gwCliQSize++;
0x00001A90 E59F1174 LDR R1,[PC,#0x0174]
0x00001A94 E5911000 LDR R1,[R1]
0x00001A98 E2811001 ADD R1,R1,#0x00000001
0x00001A9C E59F2168 LDR R2,[PC,#0x0168]
0x00001AA0 E5821000 STR R1,[R2]
28: }
29:
30:
31: static void chCheck(CHAR cTmpChar)
32: {
0x00001AA4 E12FFF1E BX R14
0),1): that gwCliQSize and the array are shared between ISR and main loop.
2) gwCliQSize is defined as volatile
3) from the assembly, bTestCh is R4 (moved from R0) and bTmpCh is R1 (moved to R0 before B)
4) I am using the J-LINK, but without J-LINK (run from flash), it still exists.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
This is inspired by/taken from this thread: http://www.allegro.cc/forums/thread/603383
The Problem
Assume the user gives you a numeric input ranging from 1 to 7. Input should be taken from the console, arguments are less desirable.
When the input is 1, print the following:
***********
*********
*******
*****
***
*
Values greater than one should generate multiples of the pattern, ending with the one above, but stacked symmetrically. For example, 3 should print the following:
*********** *********** ***********
********* ********* *********
******* ******* *******
***** ***** *****
*** *** ***
* * *
*********** ***********
********* *********
******* *******
***** *****
*** ***
* *
***********
*********
*******
*****
***
*
Bonus points if you print the reverse as well.
*********** ***********
********* *********
******* *******
***** *****
*** ***
* *
***********
*********
*******
*****
***
*
*
***
*****
*******
*********
***********
* *
*** ***
***** *****
******* *******
********* *********
*********** ***********
Can we try and keep it to one answer per language, that we all improve on?
Assembler, 165 bytes assembled
Build Instructions
Download A86 from here
Add a reference to the A86 executable into your DOS search path
Paste the code below into a text file (example: triforce.asm)
Invoke the assembler: a86 triforce.asm
This will create a .COM file called triforce.com
Type triforce to run
This was developed using the standard WinXP DOS box (Start->Programs->Accessories->Command Prompt). It should work with other DOS emulators.
Assemble using A86 and requires WinXP DOS box to run the .COM file it produces. Press 'q' to exit, keys 1-7 to draw the output.
l20:mov ah,7
int 21h
cmp al,'q'
je ret
sub al,'0'
cmp al,1
jb l20
cmp al,7
ja l20
mov [l0-1],al
mov byte ptr [l7+2],6
jmp $+2
mov ah,2
mov ch,0
mov bh,3
l0:mov bl,1
l1:mov dh,0
l3:cmp dh,ch
je l2
mov dl,32
int 21h
inc dh
jmp l3
ret
l2:mov dh,bh
l6:mov cl,12
l5:mov dl,42
cmp cl,bl
ja l4
mov dl,32
cmp dh,1
je l21
l4:int 21h
dec cl
jnz l5
l21:dec dh
jnz l6
mov dl,10
int 21h
mov dl,13
int 21h
l10:inc ch
l9:add bl,2
l7:cmp ch,6
jne l1
l13:add byte ptr [l7+2],6
l11:dec bh
l12:cmp bh,0
jne l0
xor byte ptr [l0+1],10
xor byte ptr [l9+1],40
xor byte ptr [l10+1],8
xor byte ptr [l13+1],40
sub byte ptr [l7+2],12
mov dh,[l0-1]
inc dh
xor [l12+2],dh
xor byte ptr [l11+1],8
xor byte ptr [l1+1],1
inc bh
cmp byte ptr [l0+1],11
je l0
jmp l20
It uses lots of self-modifying code to do the triforce and its mirror, it even modifies the self-modifying code.
GolfScript - 43 chars
~:!6*,{:^' '
*'*'12*' '
^6%.+)*+
-12>!^
6/-*
n}
/
~:!6*,{:^' '*'*'12*' '^6%.+)*+-12>!^6/-*n}/
48 Chars for the bonus
~:!6*,.-1%+{
:^' '*'*'12
*' '^6%.+
)*+-12>
!^6/-
*n}
/
~:!6*,.-1%+{:^' '*'*'12*' '^6%.+)*+-12>!^6/-*n}/
Python - 77 Chars
n=input()
for k in range(6*n):print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6)
n=input()
for k in range(6*n):j=1+k%6*2;print' '*k+('*'*(12-j)+' '*j)*(n-k/6)
89 Chars for the bonus
n=input();R=range(6*n)
for k in R+R[::-1]:print' '*k+('*'*11+' '*11)[k%6*2:][:12]*(n-k/6)
114 Chars Version just using string replacements
u,v=' *';s=(v*11+u)*input()
while s.strip():print s;s=u+s.replace(*((v*2+u,u*3),(v*1+u*10,v*11))[' * 'in s])[:-2]
Unk Chars all in one statement, should work w/ 2.x and 3.x. The enumerate() is to allow the single input() to work for both places you need to use it.
print ('\n'.join('\n'.join(((' '*(6*n))+' '.join(('%s%s%s'%(' '*(5-x),'*'*(2*x+1),' '*(5-x)) for m in range(i + 1)))) for x in range(5,-1,-1)) for n, i in enumerate(range(int(input())-1,-1,-1))))
Yet Another Method
def f(n): print '\n'.join(' '*6*(n-r)+(' '*(5-l)+'*'*(l*2+1)+' '*(5-l)+' ')*r for r in xrange(1, n+1) for l in xrange(6))
f(input())
Ruby - 74 Chars
(6*n=gets.to_i).times{|k|puts' '*k+('*'*(11-(j=k%6*2))+' '*(j+1))*(n-k/6)}
COBOL - 385 Chars
$ cobc -free -x triforce.cob && echo 7| ./triforce
PROGRAM-ID.P.DATA DIVISION.WORKING-STORAGE SECTION.
1 N PIC 9.
1 M PIC 99.
1 value '0100***********'.
2 I PIC 99.
2 K PIC 99.
2 V PIC X(22).
2 W PIC X(99).
PROCEDURE DIVISION.ACCEPT N
COMPUTE M=N*6
PERFORM M TIMES
DISPLAY W(1:K)NO ADVANCING
PERFORM N TIMES
DISPLAY V(I:12)NO ADVANCING
END-PERFORM
DISPLAY ''
ADD 2 TO I
IF I = 13 MOVE 1 TO I ADD -1 TO N END-IF
ADD 1 TO K
END-PERFORM.
K could be returned to outside the group level. An initial value of zero for a numeric with no VALUE clause is compiler-implementation dependent, as is an initial value of space for an alpha-numeric field (W has been cured of this, at no extra character cost). Moving K back would save two characters. -free is compiler-dependant as well, so I'm probably being over-picky.
sed, 117 chars
s/$/76543210/
s/(.).*\1//
s/./*********** /gp
:
s/\*(\**)\*/ \1 /gp
t
:c
s/\* {11}\*/ ************/
tc
s/\* / /p
t
Usage: $ echo 7 | sed -rf this.sed
First attempt; improvements could probably be made...
Ruby 1.9 - 84 characters :
v=gets.to_i
v.times{|x|6.times{|i|puts' '*6*x+(' '*i+'*'*(11-2*i)+' '*i+' ')*(v-x)}}
Perl - 72 chars
die map$"x$_.("*"x(12-($l=1+$_%6*2)).$"x$l)x($n-int$_/6).$/,0..6*($n=<>)
78 chars
map{$l=$_%6*2;print$"x$_,("*"x(11-$l).$"x$l.$")x($n-int$_/6),$/}0..6*($n=<>)-1
87 chars
$n=<>;map{$i=int$_/6;$l=$_%6*2;print$"x$_,("*"x(11-$l).$"x$l.$")x($n-$i),$/}(0..6*$n-1)
97 chars
$n=<>;map{$i=int$_/6;$l=$_%6;print$"x(6*$i),($"x$l."*"x(11-2*$l).$"x$l.$")x($n-$i),$/}(0..6*$n-1)
108 chars
$n=<>;map{$i=int$_/6;$l=$_%6;print ""." "x(6*$i),(" "x$l."*"x(11-2*$l)." "x$l." ")x($n-$i),"\n";}(0..6*$n-1)
Powershell, 78 characters
0..(6*($n=read-host)-1)|%{" "*$_+("*"*(12-($k=1+$_%6*2))+" "*$k)*(.4+$n-$_/6)}
Bonus, 92 characters
$a=0..(6*($n=read-host)-1)|%{" "*$_+("*"*(12-($k=1+$_%6*2))+" "*$k)*(.4+$n-$_/6)}
$a
$a|sort
The output is stored in an array of strings, $a, and the reverse is created by sorting the array. We could, of course, just reverse the array, but it would be more characters to type :)
Haskell - 131 138 142 143 Chars
(⊗)=replicate
z o=[concat$(6*n+m)⊗' ':(o-n)⊗((11-m-m)⊗'*'++(1+m+m)⊗' ')|n<-[0..o-1],m<-[0..5]]
main=getLine>>=mapM_ putStrLn.z.read
This one is longer (146 148 chars) at present, but an interesting, alternate line of attack:
(⊗)=replicate
a↑b|a>b=' ';_↑_='*'
z o=[map(k↑)$concat$(6*n)⊗' ':(o-n)⊗"abcdefedcba "|n<-[0..o-1],k<-"abcdef"]
main=getLine>>=mapM_ putStrLn.z.read
FORTRAN - 97 Chars
Got rid of the #define and saved 8 bytes thanks to implict loops!
$ f95 triforce.f95 -o triforce && echo 7 | ./triforce
READ*,N
DO K=0,N*6
M=2*MOD(K,6)
PRINT*,(' ',I=1,K),(('*',I=M,10),(' ',I=0,M),J=K/6+1,N)
ENDDO
END
125 bytes for the bonus
READ*,N
DO L=1,N*12
K=L+5
If(L>N*6)K=N*12-L+6
M=2*MOD(K,6)
PRINT"(99A)",(32,I=7,K),((42,I=M,10),(32,I=0,M),J=K/6,N)
ENDDO
END
FORTRAN - 108 Chars
#define R REPEAT
READ*,N
DO I=0,6*N
J=MOD(I,6)*2
PRINT*,R(' ',I)//R(R('*',11-J)//R(' ',J+1),N-I/6)
ENDDO
END
JavaScript 1.8 - SpiderMonkey - 118 chars
N=readline()
function f(n,c)n>0?(c||' ')+f(n-1,c):''
for(i=0;i<N*6;i++)print(f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1)))
w/ bonus - 151 chars
N=readline()
function f(n,c)n>0?(c||' ')+f(n-1,c):''
function l(i)print(f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1)))
for(i=0;i<N*6;i++)l(i)
for(;i--;)l(i)
Usage: js thisfile.js
JavaScript - In Browser - 154 characters
N=prompt()
function f(n,c){return n>0?(c||' ')+f(n-1,c):''}
s='<pre>'
for(i=0;i<N*6;i++)s+=f(i)+f(N-i/6,f(11-(z=i%6*2),'*')+f(z+1))+'\n'
document.write(s)
The non-obfuscated version (before optimizations by gnarf):
var N = prompt();
var S = ' ';
function fill(c, n) {
for (ret=''; n--;)
ret += c;
return ret;
}
var str = '<pre>';
for (i=0; i<N*6; i++) {
str += fill(S, i);
for (j=0; j<N-i/6; j++)
str += fill('*', 11-i%6*2) + fill(S, i%6*2+1);
str += '\n';
}
document.write(str);
Here's a different algorithm that uses replace() to go from one line to the next of each line of a triangle row:
161 characters
N=readline()
function f(n,c){return n>0?(c||' ')+f(n-1,c):''}l=0
for(i=N;i>0;){r=f(i--,f(11,'*')+' ');for(j=6;j--;){print(f(l++)+r)
r=r.replace(/\*\* /g,' ')}}
F#, 184 181 167 151 147 143 142 133 chars
let N,r=int(stdin.ReadLine()),String.replicate
for l in[0..N*6-1]do printfn"%s%s"(r l" ")(r(N-l/6)((r(11-l%6*2)"*")+(r(l%6*2+1)" ")))
Bonus, 215 212 198 166 162 158 157 148 chars
let N,r=int(stdin.ReadLine()),String.replicate
for l in[0..N*6-1]#[N*6-1..-1..0]do printfn"%s%s"(r l" ")(r(N-l/6)((r(11-l%6*2)"*")+(r(l%6*2+1)" ")))
C - 120 Chars
main(w,i,x,y){w=getchar()%8*12;for(i=0;i<w*w/2;)y=i/w,x=i++%w,putchar(x>w-2?10:x<y|w-x-1<y|(x-y)%12>=11-2*(y%6)?32:42);}
Note that this solution prints some trailing spaces (which is okay, right?). It also relies on relational operators having higher precedence than bitwise OR, saving two characters.
124 Chars
main(n,i,k){n=getchar()&7;for(k=0;k<6*n;k++,putchar(10))for(i=-k-1;++i<12*n-2*k-1;putchar(32+10*(i>=0&&(11-i%12>2*k%12))));}
C - 177 183 Chars
#define P(I,C)for(m=0;m<I;m++)putchar(C)
main(t,c,r,o,m){scanf("%d",&t);for(c=t;c>0;c--)for(r=6;r>0;r--){P((t-c)*6+6-r,32);for(o=0;o<c;o++){P(r*2-1,42);P(13-r*2,32);}puts("");}}
C - 222 243 Chars (With Bonus Points)
#define P(I,C)for(m=0;m<I;m++)putchar(C)
main(t,c,r,o,m){scanf("%d",&t);for(c=t-1;-c<2+t;c-=1+!c)for(r=c<0?1:6;c<0?r<7:r>0;r+=c<0?1:-1){P((t-abs(c+1))*6+6-r,32);for(o=0;o<abs(c+1);o++){P(r*2-1,42);P(13-r*2,32);}puts("");}}
This is my first Code Golf submission as well!
Written in C
Bonus points (492 chars):
p(char *t, int c, int s){int i=0;for(;i<s;i++)printf(" ");for(i=0;i<c;i++)printf("%s",t);printf("\n");}main(int a, char **v){int i=0;int k;int c=atoi(v[1]);for(;i<c;i++){p("*********** ",c-i,i);p(" ********* ",c-i,i);p(" ******* ",c-i,i);p(" ***** ",c-i,i);p(" *** ",c-i,i);p(" * ",c-i,i);}for(i=0;i<c;i++){k=c-i-1;p(" * ",1+i,k);p(" *** ",1+i,k);p(" ***** ",1+i,k);p(" ******* ",1+i,k);p(" ********* ",1+i,k);p("*********** ",i+1,k);}}
Without bonus points (322 chars):
p(char *t, int c, int s){int i=0;for(;i<s;i++)printf(" ");for(i=0;i<c;i++)printf("%s",t);printf("\n");}main(int a, char **v){int i=0;int k;int c=atoi(v[1]);for(;i<c;i++){p("*********** ",c-i,i);p(" ********* ",c-i,i);p(" ******* ",c-i,i);p(" ***** ",c-i,i);p(" *** ",c-i,i);p(" * ",c-i,i);}}
First time posting, too!
Lua, 121 chars
R,N,S=string.rep,io.read'*n',' 'for i=0,N-1 do for j=0,5 do X=R(S,j)print(R(S,6*i)..R(X..R('*',11-2*j)..X..S,N-i))end end
123
R,N,S=string.rep,io.read'*n',' 'for i=0,N-1 do for j=0,5 do print(R(S,6*i)..R(R(S,j)..R('*',11-2*j)..R(S,j)..S,N-i))end end
PHP, 153
<?php $i=fgets(STDIN);function r($n,$c=' '){return$n>0?$c.r($n-1,$c):'';}for($l=0;$l<$i*6;){$z=$l%6*2;echo r($l).r($i-$l++/6,r(11-$z,'*').r($z+1))."\n";}
with Bonus, 210
<?php $i=fgets(STDIN);function r($n,$c=' '){return$n>0?$c.r($n-1,$c):'';}$o=array();for($l=0;$l<$i*6;){$z=$l%6*2;$o[]=r($l).r($i-$l++/6,r(11-$z,'*').r($z+1));}print join("\n",array_merge($o,array_reverse($o)));
dc 105 chars
123 129 132 139 141
[rdPr1-d0<P]sP?sn
0sk[1lk6%2*+sj32lkd0<Plnlk6/-si
[[*]12lj-d0<P32ljd0<Pli1-dsi0<I]dsIx
10Plk1+dskln6*>K]dsKx
Mathematica, 46 characters
The answer prints sideways.
TableForm#{Table["*",{l,#},{l},{j,6},{2j-1}]}&
HyperTalk - 272 chars
function triforce n
put"******" into a
put n*6 into h
repeat with y=0 to h-1
put" " after s
put char 1 to y of s after t
repeat n-y div 6
get y mod 6*2
put char 1 to 11-it of (a&a)&&char 1 to it of s after t
end repeat
put return after t
end repeat
return t
end triforce
Indentation is neither needed nor counted (HyperCard automatically adds it).
Miscellanea:
Since there is no notion of console or way to access console arguments in HyperCard 2.2 (that I know of), a function is given instead. It can be invoked with:
on mouseUp
ask "Triforce: "
put triforce(it) into card field 1
end mouseUp
To use this, a card field would be created and set to a fixed-width font. Using HyperCard's answer command would display a dialog with the text, but it doesn't work because:
The answer dialog font (Chicago) is not fixed-width.
The answer command refuses to display long text (even triforce(2) is too long).
Common Lisp, 150 characters:
(defun f(n o)(unless(= n 0)(dotimes(x 6)(format t"~v#{~a~:*~}~-1:*~v#{~?~2:*~}~%"
o" "n"~11#: "(list(- 11(* 2 x))#\*)))(f(1- n)(+ 6 o))))
77 char alternative python solution based on gnibbler's:
n=input()
k=0
exec"print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6);k+=1;"*6*n
Amazingly the bonus came out exactly the same also (101 chars, oh well)
n=input()
l=1
k=0
s="print' '*k+('*'*12+' '*(k%6*2+1))[-12:]*(n-k/6);k+=l;"*6*n
exec s+'l=-1;k-=1;'+s