i have two char buffers which i am trying to compare parts of them. i am having a weird problem. i have the following code:
char buffer1[50], buffer2[60];
// Get buffer1 and buffer2 from the network by reading sockets
for(int i = 0; i < 20; i++)
{
if(buffer1[15+i] != buffer2[25+i])
{
printf("%c", buffer1[15+i]);
printf("%c", buffer2[25+i]);
printf("%02x", (unsigned char)buffer1[15+i]);
printf("%02x", (unsigned char)buffer2[25+i]);
break;
}
}
The above code is a simplified version of my actual code which I didnt copy-paste here because its too long. Just in case this might help, I got those two buffer over the network by reading sockets.
The problem is the loop breaks even when both the buffers are the same. To check what is in the buffers, I added the two print statements inside the if statement. And the weird thing is is, the printf statements both print the same value for %c and %02x, but the comparison fails and the loop breaks.
(Disclaimer: I'm not a C/++ expert)
It seems to me like the data is changing while you're looking at it. Two quick questions come to mind:
If you run this in the debugger, and go over the loop step-by-step, does it still happen? If it doesn't, then I strongly suspect my second question will lead you to the answer.
Is the read operation asynchronous? It seems like data is still being read while you're inside the for loop, meaning you didn't wait for the read to finish.
The only thing I see is a timing issue. If they are not the same on the if statement and they are the same on the print statement someone changed them in between.
Related
My application has crashed in an assert, and the debugger is attached. To be able to reproduce the crash I want to save a C++ vector with 397 struct{uint64_t, uint64_t} elements to file.
My first approach was to try to print the vector. I can print the vector to the console, but it seems like only the first 256 values are written. Is it possible to remove the 256 element restriction?
I've also searched for a way to save the vector to file from within the debugger, but I've not found any way. I've not even found a way to save a memory region, but I guess that must be possible...
Since you mentioned that you're stopped in the debugger in Xcode, I'll assume you're debugging with lldb. You can use the expression command to execute essentially arbitrary code when you're stopped in the debugger, for example:
expression for(int j = 0; j < 10; j++) { (void)NSLog(#"%d", j); }
Will execute a for loop and print the numbers 0 through 9. You should be able to use a similar technique to iterate over your vector and write it to a file. You can combine multiple expressions using a semicolon, just as if you were writing normal code (well, except for newlines). For example, this will write "Hello, world" to a temporary file at /tmp/vector.dat, not exactly what you want, but I think you'll get the idea:
expression FILE *fp = (FILE*)fopen("/tmp/vector.dat", "w"); (void)fprintf(fp, "Hello, world!\n"); (void)fclose(fp);
I just saw in the new OpenCV 2.4.3 that they added a universal parallel_for. So following this example, I tried to implement it myself. I got it all functioning with my code, but when I timed its processing vs a similar loop done in a typical serial fashion with a regular "for" command, the results were insignificantly faster, or often a tiny bit slower!
I thought maybe this had something to do with my pushing into vectors or something (I'm a pretty big noob to parallel processing), so I set up a test loop of just running through a big number and it still doesn't work.
Code:
class Parallel_Test : public cv::ParallelLoopBody
{
private:
double* const mypointer;
public:
Parallel_Test(double* pointer)
: mypointer(pointer){
}
void operator() (const Range& range) const
{
//This constructor needs to be here otherwise it is considered an abstract class.
// qDebug()<<"This should never be called";
}
void operator ()(const cv::BlockedRange& range) const
{
for (int x = range.begin(); x < range.end(); ++x){
mypointer[x]=x;
}
}
};
//TODO Loop pixels in parallel
double t = (double)getTickCount();
//TEST PARALELL LOOPING AT ALL
double data1[1000000];
cv::parallel_for(BlockedRange(0, 1000000), Parallel_Test(data1));
t = ((double)getTickCount() - t)/getTickFrequency();
qDebug() << "Parallel TEST time " << t << endl;
t = (double)getTickCount();
for(int i =0; i<1000000; i++){
data1[i]=i;
}
t = ((double)getTickCount() - t)/getTickFrequency();
qDebug() << "SERIAL Scan time " << t << endl;
output:
Parallel TEST time 0.00415479
SERIAL Scan time 0.00204597
Wow! I found the answer! "parallel_for" and "parallel_for_" (with a trailing underscore!) are totally different. You need the trailing underscore to make it work! Otherwise it will just run your loop in serial and you will have to use a BLOCKEDRANGE instead of a range! AHH!
Thanks to #Daniil Osokin and especially #Vladislav Vinogradov for pointing this out!
So again you code will need to look something like this:
cv::parallel_for_(Range(0, 1000000), Parallel_Test(data1));
More updated details at: http://answers.opencv.org/question/3730/how-to-use-parallel_for/
The problem is most likely that your loop body is too small.
It appears all you are doing is assigning a pointer in one vector to another.
You really need to think of a parallel for as an inefficient for loop, that is the work inside each iteration needs to be large enough so that you wouldn't dream of getting speedups by unrolling the loop because in addition to the usual decrement, compare and jump that can go on you also have a few interlocked instructions and perhaps a virtual function call or two and some allocations.
So instead of copying a pointer try doing a good amount of real math or work on a large array of data.
I'm having trouble here. I launch two kernels , check if some value is the one expected (memcpy to the host), if it is I stop, if it isn't I launch the two kernels again.
the first kernel:
__global__ void aco_step(const KPDeviceData* data)
{
int obj = threadIdx.x;
int ant = blockIdx.x;
int id = threadIdx.x + blockIdx.x * blockDim.x;
*(data->added) = 1;
while(*(data->added) == 1)
{
*(data->added) = 0;
//check if obj fits
int fits = (data->obj_weights[obj] + data->weight[ant] <= data->max_weight);
fits = fits * !(getElement(data->selections, data->selections_pitch, ant, obj));
if(obj == 0)
printf("ant %d going..\n", ant);
__syncthreads();
...
The code goes on after this. But that printf never gets printed, that syncthreads is there just for debugging purposes.
The "added" variable was shared, but since shared memory is a PITA and usually throws bugs in the code, i just removed it for now. This "added" variable isn't the smartest thing to do but it's faster than the alternative, which is checking if any variable within an array is some value on the host and deciding to keep iterating or not.
The getElement, simply does the matrix memory calculation with the pitch to access the right position and returns the element there:
int* el = (int*) ((char*)mat + row * pitch) + col;
return *el;
The obj_weights array has the right size, n*sizeof(int). So does the weight array, ants*sizeof(float). So they aren't out of bounds.
The kernel after this one has a printf right on the beginning, and it doesn't get printed either and after the printf it sets a variable on the device memory, and this memory is copied to the CPU after the kernel finished, and it isn't the right value when I print it in the CPU code. So I think this kernel is doing something illegal and the second one doesn't even get launched.
I'm testing some instances, when I launch 8 blocks and 512 threads, it runs OK. 32 blocks, 512 threads, OK. But 8 blocks and 1024 threads, and this happens, the kernel doesn't work, neither 32 blocks and 1024 threads.
Am I doing something wrong? Memory access? Am I launching too many threads?
edit: tried removing the "added" variable and the while loop, so it should execute just once. Still doesnt work, nothing gets printed, even if the printf is right after the three initial lines and the next kernel also doesn't print anything.
edit: another thing, I'm using a GTX 570, so the "Maximum number of threads per block" is 1024 according to http://en.wikipedia.org/wiki/CUDA. Maybe I'll just stick with 512 maximum or check on how higher I can put this value.
__syncthreads() inside conditional code is only allowed if the condition evaluates identically on all threads of a block.
In your case the condition suffers a race condition and is nondeterministic, so it most probably evaluates to different results for different threads.
printf() output is only displayed after the kernel finishes successfully. In this case it doesn't due to the problem mentioned above, so the output never shows up. You could have figured out this by testing the return codes all CUDA function calls for errors.
I am trying to emulate grep pattern of UNIX using a C program( just for learning ). The code that i have written is giving me a run time error..
#include <fcntl.h>
#include <stdio.h>
#include <string.h>
#define MAXLENGTH 1000
char userBuf[MAXLENGTH];
int main ( int argc, char *argv[])
{
int numOfBytes,fd,i;
if (argc != 2)
printf("Supply correct number of arguments.\n");
//exit(1);
fd =open("pattern.txt",O_RDWR);
if ( fd == -1 )
printf("File does not exist.\n");
//exit(1);
while ( (numOfBytes = read(fd,userBuf,MAXLENGTH)) > 0 )
;
printf("NumOfBytes = %d\n",numOfBytes);
for(i=0;userBuf[i] != '\0'; ++i)
{
if ( strstr(userBuf,argv[1]) )
printf("%s\n",userBuf);
}
}
The program is printing infinitely, the lines containing the pattern . I tried debugging , but couldn't figure out the error. Please let me know where am i wrong.,
Thanks
Say the string is "fooPATTERN". Your first time through the loop, you check for the pattern in "fooPATTERN" and find it. Then your second time through the loop, you check for the pattern in "ooPATTERN" and find it again. Then your third time, you check for the pattern in "oPATTERN" and find it again.
Since you're doing this to learn, I won't tell you much more. You can decide how best to solve it. There are at least two fundamentally different ways you could solve it. One is to do less on each pass of the loop to ensure you only find it once. The other is to make sure your next pass of the loop is past any pattern that was found.
One thing to think about: If the pattern is 'oo' and the string is 'ooo', how many patterns should be found? 1 or 2?
The 'read' does not delimit the data with a null character.
The while loop should encompase the for loop - it doesn't
First, you shouldn't be using raw Unix i/o with open and read if you're just learning C. Start with standard C i/o with fopen and fread/fscanf/fgets and so forth.
Second, you're reading in successive pieces of the file into the same buffer, overwriting the buffer each time, and only ever processing the last contents of the buffer.
Third, nothing guarantees that your buffer will be zero-terminated when you read into it with read(). In fact, it usually won't be.
Fourth, you're not using the i variable in the body of your loop. I can't tell exactly what you were shooting for here, but doing the same thing on the same data umpteen thousand times surely wasn't it.
Fifth, always compile with the fullest warning settings you can abide -- at lest -Wall with GCC. It should have complained that you call read() without including <unistd.h>.
in a simple MPI program I have used a column wise division of a large matrix.
How can I order the output so that each matrix appears next to the other ordered ?
I have tried this simple code the effect is quite different from the wanted:
for(int i=0;i<10;i++)
{
for(int k=0;k<numprocs;k++)
{
if (my_id==k){
for(int j=1;j<10;j++)
printf("%d",data[i][j]);
}
MPI_Barrier(com);
}
if(my_id==0)
printf("\n");
}
Seems that each process has his own stdout and so is impossible to have ordered lines output without sending all the data to one master which will print out. Is my guess true ? Or what I'm doing wrong ?
You guessed right. The MPI standard does not specify how stdout from different nodes should be collected for printing at the originating process. It is often the case that when multiple processes are doing prints the output will get merged in an unspecified way. fflush doesn't help.
If you want the output ordered in a certain way, the most portable method would be to send the data to the master process for printing.
For example, in pseudocode:
if (rank == 0) {
print_col(0);
for (i = 1; i < comm_size; i++) {
MPI_Recv(buffer, .... i, ...);
print_col(i);
}
} else {
MPI_Send(data, ..., 0, ...);
}
Another method which can sometimes work would be to use barries to lock step processes so that each process prints in turn. This of course depends on the MPI Implementation and how it handles stdout.
for(i = 0; i < comm_size; i++) {
MPI_Barrier(MPI_COMM_WORLD);
if (i == rank) {
printf(...);
}
}
Of course, in production code where the data is too large to print sensibly anyway, data is eventually combine by having each process writing to a separate file and merged separately, or using MPI I/O (defined in the MPI2 standards) to coordinate parallel writes.
I produced ordered output to a file before using the exact same method. You could try printing to a temporary file, printing the contents of said file and then deleting it.
Have the root processor do all of the printing. Use MPI_Send/MPI_Recv or MPI_Gather (or whatever) to send the data in turn from each processor to the root.
To solve this problem you can use short sleep. I use and then it works in 99%
printf("text nr 1\n");
MPI_Barrier(MPI_COMM_WORLD);
usleep(100);
printf("text nr 2\n");
It's not very elegant but works.