I am having trouble using cudaMemset on a device variable. Is it possible to use the reference to the device variable for cudaMemset, or is it just a matter of missing compiler flags, or libraries.. I am using cuda 4.1, and
NVRM version: NVIDIA UNIX x86_64 Kernel Module 285.05.33 Thu Jan 19
14:07:02 PST 2012
This is my sample code:
#include <stdio.h>
#include <stdlib.h>
#include <cuda_runtime.h>
// device variable and kernel
__device__ float d_test;
int main() {
if (cudaMemset(&d_test,0,sizeof(float)) !=cudaSuccess)
printf("Error!\n");
}
which outputs:
Error!
Your problem is that d_test (as it appears in the host symbol table) isn't a valid device address and the runtime cannot access it directly. The solution is to use the cudaGetSymbolAddress API function to read the address of the device symbol from the context at runtime. Here is a slightly expanded version of your demonstration case which should work correctly:
#include <stdio.h>
#include <stdlib.h>
#include <cuda_runtime.h>
// device variable and kernel
__device__ float d_test;
inline void gpuAssert(cudaError_t code, char * file, int line, bool Abort=true)
{
if (code != cudaSuccess) {
fprintf(stderr, "GPUassert: %s %s %d\n", cudaGetErrorString(code),file,line);
if (Abort) exit(code);
}
}
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
int main()
{
float * _d_test;
gpuErrchk( cudaFree(0) );
gpuErrchk( cudaGetSymbolAddress((void **)&_d_test, "d_test") );
gpuErrchk( cudaMemset(_d_test,0,sizeof(float)) );
gpuErrchk( cudaThreadExit() );
return 0;
}
Here, we read the address of the device symbol d_test from the context into a host pointer _d_test. This can then be passed to host side API functions like cudaMemset, cudaMemcpy, etc.
Edit to note that the form of cudaGetSymbolAddress shown in this answer has been deprecated and removed from the CUDA runtime API. For modern CUDA, the call would be:
gpuErrchk( cudaGetSymbolAddress((void **)&_d_test, d_test) );
I believe you can also use cudaMemcpyFromSymbol:
A function, such as the following kernel, can change the value of the variable declared in global memory (outside of the main function)
__global__ void kernel1() { d_test = 1.0; }
Inside your main, you can obtain the value using cudaMemcpyFromSymbol
cudaMemcpyFromSymbol(&h_test,"d_test",sizeof(float),0,cudaMemcpyDeviceToHost);
Of course, there is also cudaMemcpyToSymbol to change the value of the global variable.
The idea came from here: Having problem assigning a device variable in CUDA
Related
I need to make my kernel communicate with the host. I tried to use a global counter (better ways are well accepted), but the following code prints always 0. What am I doing wrong? (I tried both commented and uncommented ways).
#include <stdio.h>
#include <cuda_runtime.h>
//__device__ int count[1] = {0};
__device__ int count = 0;
__global__ void inc() {
//count[0]++;
atomicAdd(&count, 1);
}
int main(void) {
inc<<<1,10>>>();
cudaDeviceSynchronize();
//int *c;
int c;
cudaMemcpyFromSymbol(&c, count, sizeof(int), cudaMemcpyDeviceToHost);
printf("%d\n", c);
return 0;
}
Anytime you are having trouble with a CUDA code, I strongly encourage you to use proper CUDA error checking and run your code with cuda-memcheck, before asking others for help. Even if you don't understand the error output, providing it in your question will be useful for those trying to help you.
If you had done so, you would have received a report that cudaMemcpyFromSymbol is throwing an invalid argument error.
If you study the documentation for that function call, you will see that the 4th parameter is not the direction parameter, but is the offset parameter. So providing cudaMemcpyDeviceToHost is incorrect for the offset parameter. Since cudaMemcpyFromSymbol is always a device->host transfer, providing the direction argument is redundant, and since it is provided a default, is unnecessary. Your code works correctly for me simply by eliminating that:
$ cat t1414.cu
#include <stdio.h>
#include <cuda_runtime.h>
//__device__ int count[1] = {0};
__device__ int count = 0;
__global__ void inc() {
//count[0]++;
atomicAdd(&count, 1);
}
int main(void) {
inc<<<1,10>>>();
cudaDeviceSynchronize();
//int *c;
int c;
cudaMemcpyFromSymbol(&c, count, sizeof(int));
printf("%d\n", c);
return 0;
}
$ nvcc -o t1414 t1414.cu
$ cuda-memcheck ./t1414
========= CUDA-MEMCHECK
10
========= ERROR SUMMARY: 0 errors
$
I'm writing a CUDA kernel that is compiled at runtime using NVRTC (CUDA version 9.2 with NVRTC version 7.5), which needs the stdint.h header, in order to have the int32_t etc. types.
If I write the kernel source code without the include, it works correctly. For example the kernel
extern "C" __global__ void f() { ... }
Compiles to PTX code where f is defined as .visible .entry f.
But if the kernel source code is
#include <stdint.h>
extern "C" __global__ void f() { ... }
it reports A function without execution space annotations (__host__/__device__/__global__) is considered a host function, and host functions are not allowed in JIT mode. (also without extern "C").
Passing -default-device makes the PTX code .visible .func f, so the function cannot be called from the host.
Is there a way to include headers in the source code, and still have a __global__ entry function? Or alternately, a way to know which integer size convention is used on the by the NVRTC compiler, so that the int32_t etc. types can be manually defined?
Edit:
Example program that shows the problem:
#include <cstdlib>
#include <string>
#include <vector>
#include <memory>
#include <cassert>
#include <iostream>
#include <cuda.h>
#include <cuda_runtime.h>
#include <nvrtc.h>
[[noreturn]] void fail(const std::string& msg, int code) {
std::cerr << "error: " << msg << " (" << code << ')' << std::endl;
std::exit(EXIT_FAILURE);
}
std::unique_ptr<char[]> compile_to_ptx(const char* program_source) {
nvrtcResult rv;
// create nvrtc program
nvrtcProgram prog;
rv = nvrtcCreateProgram(
&prog,
program_source,
"program.cu",
0,
nullptr,
nullptr
);
if(rv != NVRTC_SUCCESS) fail("nvrtcCreateProgram", rv);
// compile nvrtc program
std::vector<const char*> options = {
"--gpu-architecture=compute_30"
};
//options.push_back("-default-device");
rv = nvrtcCompileProgram(prog, options.size(), options.data());
if(rv != NVRTC_SUCCESS) {
std::size_t log_size;
rv = nvrtcGetProgramLogSize(prog, &log_size);
if(rv != NVRTC_SUCCESS) fail("nvrtcGetProgramLogSize", rv);
auto log = std::make_unique<char[]>(log_size);
rv = nvrtcGetProgramLog(prog, log.get());
if(rv != NVRTC_SUCCESS) fail("nvrtcGetProgramLog", rv);
assert(log[log_size - 1] == '\0');
std::cerr << "Compile error; log:\n" << log.get() << std::endl;
fail("nvrtcCompileProgram", rv);
}
// get ptx code
std::size_t ptx_size;
rv = nvrtcGetPTXSize(prog, &ptx_size);
if(rv != NVRTC_SUCCESS) fail("nvrtcGetPTXSize", rv);
auto ptx = std::make_unique<char[]>(ptx_size);
rv = nvrtcGetPTX(prog, ptx.get());
if(rv != NVRTC_SUCCESS) fail("nvrtcGetPTX", rv);
assert(ptx[ptx_size - 1] == '\0');
nvrtcDestroyProgram(&prog);
return ptx;
}
const char program_source[] = R"%%%(
//#include <stdint.h>
extern "C" __global__ void f(int* in, int* out) {
out[threadIdx.x] = in[threadIdx.x];
}
)%%%";
int main() {
CUresult rv;
// initialize CUDA
rv = cuInit(0);
if(rv != CUDA_SUCCESS) fail("cuInit", rv);
// compile program to ptx
auto ptx = compile_to_ptx(program_source);
std::cout << "PTX code:\n" << ptx.get() << std::endl;
}
When //#include <stdint.h> in the kernel source is uncommented it no longer compiles. When //options.push_back("-default-device"); is uncommented it compiles but does not mark the function f as .entry.
CMakeLists.txt to compile it (needs CUDA driver API + NVRTC)
cmake_minimum_required(VERSION 3.4)
project(cudabug CXX)
find_package(CUDA REQUIRED)
set(CMAKE_CXX_STANDARD 14)
set(CMAKE_CXX_STANDARD_REQUIRED 14)
add_executable(cudabug cudabug.cc)
include_directories(SYSTEM ${CUDA_INCLUDE_DIRS})
link_directories(${CUDA_LIBRARY_DIRS})
target_link_libraries(cudabug PUBLIC ${CUDA_LIBRARIES} nvrtc cuda)
[Preface: this is a very hacky answer, and is specific to the GNU toolchain (although I suspect the problem in the question is also specific to the GNU toolchain)].
It would appear that the problem here is with the GNU standard header features.h, which gets pulled into stdint.h and then winds up defining a lot of stub functions which have the default __host__ compilation space. This causes nvrtc to blow up. It also seems that the -default-device option will result in a resolved glibC compiler feature set which makes the whole nvrtc compiler fail.
You can defeat this (in a very hacky way) by predefining a feature set for the standard library which excludes all the host functions. Changing your JIT kernel code to
const char program_source[] = R"%%%(
#define __ASSEMBLER__
#define __extension__
#include <stdint.h>
extern "C" __global__ void f(int32_t* in, int32_t* out) {
out[threadIdx.x] = in[threadIdx.x];
}
)%%%";
got me this:
$ nvcc -std=c++14 -ccbin=g++-7 jit_header.cu -o jitheader -lnvrtc -lcuda
$ ./jitheader
PTX code:
//
// Generated by NVIDIA NVVM Compiler
//
// Compiler Build ID: CL-24330188
// Cuda compilation tools, release 9.2, V9.2.148
// Based on LLVM 3.4svn
//
.version 6.2
.target sm_30
.address_size 64
// .globl f
.visible .entry f(
.param .u64 f_param_0,
.param .u64 f_param_1
)
{
.reg .b32 %r<3>;
.reg .b64 %rd<8>;
ld.param.u64 %rd1, [f_param_0];
ld.param.u64 %rd2, [f_param_1];
cvta.to.global.u64 %rd3, %rd2;
cvta.to.global.u64 %rd4, %rd1;
mov.u32 %r1, %tid.x;
mul.wide.u32 %rd5, %r1, 4;
add.s64 %rd6, %rd4, %rd5;
ld.global.u32 %r2, [%rd6];
add.s64 %rd7, %rd3, %rd5;
st.global.u32 [%rd7], %r2;
ret;
}
Big caveat: This worked on the glibC system I tried it on. It probably won't work with other toolchains or libC implementations (if, indeed, they have this problem).
Another alternative is creating stand-ins, for some of the standard library headers. NVRTC's API supports your specifying header file contents as strings, associated with header names - before it will go looking through the filesystem for you. This approach is adopted in NVIDIA JITify, and I've adopted it myself working on something else which may or may not be released.
The easy way to do this You can just take the JITify header stubs for stdint.h, limits.h , from here, which I'm also attaching since it's not very long. Alternatively, you can generate this stub yourself to make sure you're not missing out on anything that's relevant from the standard. Here's how that works:
Start with your stdint.h file (or cstdint file as the case may be);
For each include directive in the file (and recursively, for each include in an include etc):
2.1 Figure out whether you can skip including the file altogether (possibly by making a few defines which are known to hold on the GPU).
2.2 If you're not sure you can skip the file - include it entirely and recurse to (2.), or keep it as its own separate header (and apply the whole process in (1.) to it).
You now have a header file which only includes device-safe header files (or none at all)
Partially-preprocess the file, dropping everything that won't be used on a GPU
Remove the lines which might be problematic on a GPU (e.g. #pragma's), and add __device__ __host__ or just __host__ as appropriate to each function declaration.
Important note: Doing this requires paying attention to licenses and copyrights. You would be creating a "derivative work" of glibc and/or JITify and/or StackOverflow contributions etc.
Now, the stdint.h and limits.h from NVIDIA JITify I promised. I've adapted them to not have namespaces:
stdint.h:
#pragma once
#include <limits.h>
typedef signed char int8_t;
typedef signed short int16_t;
typedef signed int int32_t;
typedef signed long long int64_t;
typedef signed char int_fast8_t;
typedef signed short int_fast16_t;
typedef signed int int_fast32_t;
typedef signed long long int_fast64_t;
typedef signed char int_least8_t;
typedef signed short int_least16_t;
typedef signed int int_least32_t;
typedef signed long long int_least64_t;
typedef signed long long intmax_t;
typedef signed long intptr_t; //optional
typedef unsigned char uint8_t;
typedef unsigned short uint16_t;
typedef unsigned int uint32_t;
typedef unsigned long long uint64_t;
typedef unsigned char uint_fast8_t;
typedef unsigned short uint_fast16_t;
typedef unsigned int uint_fast32_t;
typedef unsigned long long uint_fast64_t;
typedef unsigned char uint_least8_t;
typedef unsigned short uint_least16_t;
typedef unsigned int uint_least32_t;
typedef unsigned long long uint_least64_t;
typedef unsigned long long uintmax_t;
#define INT8_MIN SCHAR_MIN
#define INT16_MIN SHRT_MIN
#if defined _WIN32 || defined _WIN64
#define WCHAR_MIN SHRT_MIN
#define WCHAR_MAX SHRT_MAX
typedef unsigned long long uintptr_t; //optional
#else
#define WCHAR_MIN INT_MIN
#define WCHAR_MAX INT_MAX
typedef unsigned long uintptr_t; //optional
#endif
#define INT32_MIN INT_MIN
#define INT64_MIN LLONG_MIN
#define INT8_MAX SCHAR_MAX
#define INT16_MAX SHRT_MAX
#define INT32_MAX INT_MAX
#define INT64_MAX LLONG_MAX
#define UINT8_MAX UCHAR_MAX
#define UINT16_MAX USHRT_MAX
#define UINT32_MAX UINT_MAX
#define UINT64_MAX ULLONG_MAX
#define INTPTR_MIN LONG_MIN
#define INTMAX_MIN LLONG_MIN
#define INTPTR_MAX LONG_MAX
#define INTMAX_MAX LLONG_MAX
#define UINTPTR_MAX ULONG_MAX
#define UINTMAX_MAX ULLONG_MAX
#define PTRDIFF_MIN INTPTR_MIN
#define PTRDIFF_MAX INTPTR_MAX
#define SIZE_MAX UINT64_MAX
limits.h:
#pragma once
#if defined _WIN32 || defined _WIN64
#define __WORDSIZE 32
#else
#if defined __x86_64__ && !defined __ILP32__
#define __WORDSIZE 64
#else
#define __WORDSIZE 32
#endif
#endif
#define MB_LEN_MAX 16
#define CHAR_BIT 8
#define SCHAR_MIN (-128)
#define SCHAR_MAX 127
#define UCHAR_MAX 255
enum {
_JITIFY_CHAR_IS_UNSIGNED = (char)-1 >= 0,
CHAR_MIN = _JITIFY_CHAR_IS_UNSIGNED ? 0 : SCHAR_MIN,
CHAR_MAX = _JITIFY_CHAR_IS_UNSIGNED ? UCHAR_MAX : SCHAR_MAX,
};
#define SHRT_MIN (-32768)
#define SHRT_MAX 32767
#define USHRT_MAX 65535
#define INT_MIN (-INT_MAX - 1)
#define INT_MAX 2147483647
#define UINT_MAX 4294967295U
#if __WORDSIZE == 64
# define LONG_MAX 9223372036854775807L
#else
# define LONG_MAX 2147483647L
#endif
#define LONG_MIN (-LONG_MAX - 1L)
#if __WORDSIZE == 64
#define ULONG_MAX 18446744073709551615UL
#else
#define ULONG_MAX 4294967295UL
#endif
#define LLONG_MAX 9223372036854775807LL
#define LLONG_MIN (-LLONG_MAX - 1LL)
#define ULLONG_MAX 18446744073709551615ULL
I'm working through the examples of the "CUDA by Example" book. The following code doesn't give me an answer and work as it should. Where's the mistake?
Will appreciate your help and answers.
I get an output,which reads
Calculation done on GPU yields the answer: &d
Press enter to stop
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <iostream>
#include <stdio.h>
using namespace std;
__global__ void add_integers_cuda(int a, int b, int *c)
{
*c = a + b;
}
int main(void)
{
int c;
int *dev_ptr;
cudaMalloc((void **)&dev_ptr, sizeof(int)); //allocate sizeof(int) bytes of contiguous memory in the gpu device and return the address of first byte to dev_ptr.
// call the kernel
add_integers_cuda <<<1,1>>>(2,7,dev_ptr);
cudaMemcpy(&c, dev_ptr, sizeof(int), cudaMemcpyDeviceToHost);
printf("Calculation done on GPU yields the answer: &d\n",c );
cudaFree(dev_ptr);
printf("Press enter to stop.");
cin.ignore(255, '\n');
return 0;
}
"
&d is not a correct printf formatting character here:
printf("Calculation done on GPU yields the answer: &d\n",c );
You won't get the output you are expecting.
You should use %d instead:
printf("Calculation done on GPU yields the answer: %d\n",c );
This particular issue has nothing to do with CUDA of course.
You may also want to run CUDA codes with cuda-memcheck and/or use proper CUDA error checking if you are just learning and having trouble. Neither of those would have pointed out the above error, however.
In jCuda one can load cuda files as PTX or CUBIN format and call(launch) __global__ functions (kernels) from Java.
With keeping that in mind, I want to develop a framework with JCuda that gets user's __device__ function in a .cu file at run-time, loads and runs it.
And I have already implemented a __global__ function, in which each thread finds out the start point of its related data, perform some computation, initialization and then call user's __device__ function.
Here is my kernel pseudo code:
extern "C" __device__ void userFunc(args);
extern "C" __global__ void kernel(){
// initialize
userFunc(args);
// rest of the kernel
}
And user's __device__ function:
extern "C" __device__ void userFunc(args){
// do something
}
And in Java side, here is the part that I load the modules(modules are made from ptx files which are successfully created from cuda files with this command: nvcc -m64 -ptx path/to/cudaFile -o cudaFile.ptx)
CUmodule kernelModule = new CUmodule(); // 1
CUmodule userFuncModule = new CUmodule(); // 2
cuModuleLoad(kernelModule, ptxKernelFileName); // 3
cuModuleLoad(userFuncModule, ptxUserFuncFileName); // 4
When I try to run it I got error at line 3 : CUDA_ERROR_NO_BINARY_FOR_GPU. After some searching I get that my ptx file has some syntax error. After running this suggested command:
ptxas -arch=sm_30 kernel.ptx
I got:
ptxas fatal : Unresolved extern function 'userFunc'
Even when I replace line 3 with 4 to load userFunc before kernel I get this error. I got stuck at this phase. Is this the correct way to load multiple modules that need to be linked together in JCuda? Or is it even possible?
Edit:
Second part of the question is here
The really short answer is: No, you can't load multiple modules into a context in the runtime API.
You can do what you want, but it requires explicit setup and execution of a JIT linking call. I have no idea how (or even whether) that has been implemented in JCUDA, but I can show you how to do it with the standard driver API. Hold on...
If you have a device function in one file, and a kernel in another, for example:
// test_function.cu
#include <math.h>
__device__ float mathop(float &x, float &y, float &z)
{
float res = sin(x) + cos(y) + sqrt(z);
return res;
}
and
// test_kernel.cu
extern __device__ float mathop(float & x, float & y, float & z);
__global__ void kernel(float *xvals, float * yvals, float * zvals, float *res)
{
int tid = threadIdx.x + blockIdx.x * blockDim.x;
res[tid] = mathop(xvals[tid], yvals[tid], zvals[tid]);
}
You can compile them to PTX as usual:
$ nvcc -arch=sm_30 -ptx test_function.cu
$ nvcc -arch=sm_30 -ptx test_kernel.cu
$ head -14 test_kernel.ptx
//
// Generated by NVIDIA NVVM Compiler
//
// Compiler Build ID: CL-19324607
// Cuda compilation tools, release 7.0, V7.0.27
// Based on LLVM 3.4svn
//
.version 4.2
.target sm_30
.address_size 64
// .globl _Z6kernelPfS_S_S_
.extern .func (.param .b32 func_retval0) _Z6mathopRfS_S_
At runtime, your code must create a JIT link session, add each PTX to the linker session, then finalise the linker session. This will give you a handle to a compiled cubin image which can be loaded as a module as usual. The simplest possible driver API code to put this together looks like this:
#include <cstdio>
#include <cuda.h>
#define drvErrChk(ans) { drvAssert(ans, __FILE__, __LINE__); }
inline void drvAssert(CUresult code, const char *file, int line, bool abort=true)
{
if (code != CUDA_SUCCESS) {
fprintf(stderr, "Driver API Error %04d at %s %d\n", int(code), file, line);
exit(-1);
}
}
int main()
{
cuInit(0);
CUdevice device;
drvErrChk( cuDeviceGet(&device, 0) );
CUcontext context;
drvErrChk( cuCtxCreate(&context, 0, device) );
CUlinkState state;
drvErrChk( cuLinkCreate(0, 0, 0, &state) );
drvErrChk( cuLinkAddFile(state, CU_JIT_INPUT_PTX, "test_function.ptx", 0, 0, 0) );
drvErrChk( cuLinkAddFile(state, CU_JIT_INPUT_PTX, "test_kernel.ptx" , 0, 0, 0) );
size_t sz;
char * image;
drvErrChk( cuLinkComplete(state, (void **)&image, &sz) );
CUmodule module;
drvErrChk( cuModuleLoadData(&module, image) );
drvErrChk( cuLinkDestroy(state) );
CUfunction function;
drvErrChk( cuModuleGetFunction(&function, module, "_Z6kernelPfS_S_S_") );
return 0;
}
You should be able to compile and run this as posted and verify it works OK. It should serve as a template for a JCUDA implementation, if they have JIT linking support implemented.
If I call cudaMemcpy from host memory to host memory, will it first synchronize the device? Is there any difference between the cuda memcpy call and the ordinary C++ function memcpy? I know that in case I want to do a memcpy 2D between host to host, I have to use the cuda call, since there is no such function in C++. Is there any other ones?
If I call cudaMemcpy from host memory to host memory, will it first synchronize the device?
I verified that cudaMemcpy() with cudaMemcpyHostToHost does synchronize with the following code:
#include <cuda.h>
#define check_cuda_call(ans) { _check((ans), __FILE__, __LINE__); }
inline void _check(cudaError_t code, char *file, int line)
{
if (code != cudaSuccess) {
fprintf(stderr,"CUDA Error: %s %s %d\n", cudaGetErrorString(code), file, line);
exit(code);
}
}
__device__ clock_t offset;
__global__ void clock_block(clock_t clock_count)
{
clock_t start_clock = clock();
clock_t clock_offset = 0;
while (clock_offset < clock_count) {
clock_offset = clock() - start_clock;
}
offset = clock_offset;
}
int main(int argc, char *argv[])
{
int *A;
check_cuda_call(cudaMallocHost(&A, 1 * sizeof(int)));
int *B;
check_cuda_call(cudaMallocHost(&B, 1 * sizeof(int)));
clock_block<<<1,1>>>(1000 * 1000 * 1000);
//check_cuda_call(cudaDeviceSynchronize());
check_cuda_call(cudaMemcpy(&A, &B, 1 * sizeof(int), cudaMemcpyHostToHost));
}
With a blocking call after the kernel launch, the app waits for around 1 second on my card. Without a blocking call, it exits immediately.
Is there any difference between the cuda memcpy call and the ordinary C++ function memcpy?
Yes, the synchronization, which also causes the cudaMemcpy() with cudaMemcpyHostToHost to be able to return errors from previous async calls, makes it different from plain memcpy().
I know that in case I want to do a memcpy 2D between host to host, I have to use the cuda call, since there is no such function in C++. Is there any other ones?
You might be able to use cudaMemcpyAsync() with cudaMemcpyHostToHost to do copies on the host without blocking the CPU, but I haven't tested it.