Cuda atomic lock: threads in sequence - cuda

Question

I have a code of which a section needs to be executed critically. I am using a lock for that piece of code so that each thread of the kernel (set up with one thread per block) executes that piece of code atomically. The order of the threads is what bothers me - I need the threads to execute in chronological order according to their indices (or actually, in order of their blockIdx), from 0 to say 10 (instead of randomly e.g. 5, 8, 3, 0, ...etc). Is it possible to do that?
Here is an example code:
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<math_functions.h>
#include<time.h>
#include<cuda.h>
#include<cuda_runtime.h>
// number of blocks
#define nob 10
struct Lock{
int *mutex;
Lock(void){
int state = 0;
cudaMalloc((void**) &mutex, sizeof(int));
cudaMemcpy(mutex, &state, sizeof(int), cudaMemcpyHostToDevice);
}
~Lock(void){
cudaFree(mutex);
}
__device__ void lock(void){
while(atomicCAS(mutex, 0, 1) != 0);
}
__device__ void unlock(void){
atomicExch(mutex, 0);
}
};
__global__ void theKernel(Lock myLock){
int index = blockIdx.x; //using only one thread per block
// execute some parallel code
// critical section of code (thread with index=0 needs to start, followed by index=1, etc.)
myLock.lock();
printf("Thread with index=%i inside critical section now...\n", index);
myLock.unlock();
}
int main(void)
{
Lock myLock;
theKernel<<<nob, 1>>>(myLock);
return 0;
}
which gives the following results:
Thread with index=1 inside critical section now...
Thread with index=0 inside critical section now...
Thread with index=5 inside critical section now...
Thread with index=9 inside critical section now...
Thread with index=7 inside critical section now...
Thread with index=6 inside critical section now...
Thread with index=3 inside critical section now...
Thread with index=2 inside critical section now...
Thread with index=8 inside critical section now...
Thread with index=4 inside critical section now...
I want these indices to start from 0 and execute chronologically to 9.
One way I thought to modify the Lock to achieve this is as follows:
struct Lock{
int *indexAllow;
Lock(void){
int startVal = 0;
cudaMalloc((void**) &indexAllow, sizeof(int));
cudaMemcpy(indexAllow, &startVal, sizeof(int), cudaMemcpyHostToDevice);
}
~Lock(void){
cudaFree(indexAllow);
}
__device__ void lock(int index){
while(index!=*indexAllow);
}
__device__ void unlock(void){
atomicAdd(indexAllow,1);
}
};
and then to just initialize the lock by passing the index as an argument:
myLock.lock(index);
but this stalls my pc... I'm probably missing something obvious.
If anyone can help I'd appreciate it!
Thanks!!!

Answer 1

I changed your code a bit. Now it produces your desired output:
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<math_functions.h>
#include<time.h>
#include<cuda.h>
#include<cuda_runtime.h>
// number of blocks
#define nob 10
struct Lock{
int *mutex;
Lock(void){
int state = 0;
cudaMalloc((void**) &mutex, sizeof(int));
cudaMemcpy(mutex, &state, sizeof(int), cudaMemcpyHostToDevice);
}
~Lock(void){
cudaFree(mutex);
}
__device__ void lock(uint compare){
while(atomicCAS(mutex, compare, 0xFFFFFFFF) != compare); //0xFFFFFFFF is just a very large number. The point is no block index can be this big (currently).
}
__device__ void unlock(uint val){
atomicExch(mutex, val+1);
}
};
__global__ void theKernel(Lock myLock){
int index = blockIdx.x; //using only one thread per block
// execute some parallel code
// critical section of code (thread with index=0 needs to start, followed by index=1, etc.)
myLock.lock(index);
printf("Thread with index=%i inside critical section now...\n", index);
__threadfence_system(); // For the printf. I'm not sure __threadfence_system() can guarantee the order for calls to printf().
myLock.unlock(index);
}
int main(void)
{
Lock myLock;
theKernel<<<nob, 1>>>(myLock);
return 0;
}
The lock() function accepts compare as the parameter and checks if it is equal to the value alraedy in mutex. If yes, it puts 0xFFFFFFFF into the mutex to indicate the lock is acquired by a thread. Because the mutex is initialized in the constructor by 0, only the thread with block ID 0 would be successful in acquiring the lock. In the unlock, we place the next block ID index into the mutex to guarantee your desired ordering. Also, because you have used printf() inside the CUDA kernel, I think a call to threadfence_system() is required for you to see them in the output in the same order.