How to capture output of execvp - unix

I'm developing a program which executes a program using execvp. It needs to capture the results of the child process and parse them in the main process. It seems there is a way, using named pipes, and duping. I'm trying to hunt down a good example of this, but so far no luck. If anyone has any pointers, links and/or suggestions about this, I'd greatly appreciate it.

You don't need named pipes; unnamed pipes work just fine. Actually, often you can just use popen instead of doing the pipe/fork/dup/exec yourself. popen works like this (though your libc's implementation likely has more error checking):
FILE *popen(const char *command, const char *type) {
int fds[2];
const char *argv[4] = {"/bin/sh", "-c", command};
pipe(fds);
if (fork() == 0) {
close(fds[0]);
dup2(type[0] == 'r' ? 0 : 1, fds[1]);
close(fds[1]);
execvp(argv[0], argv);
exit(-1);
}
close(fds[1]);
return fdopen(fds[0], type);
}
This creates an unnamed pipe, and forks. In the child, it reattaches stdout (or stdin) to one end of the pipe, then execs the child. The parent can simply read (or write) from the other end of the pipe.

Can't you just use popen()?

Here is a simple example that demonstrates the use of popen to achieve your goal. Just put something more interesting than "echo" as the command :)
#include <stdio.h>
int main()
{
char buf[100];
int i = 0;
FILE *p = popen("echo \"Test\"","r");
if (p != NULL )
{
while (!feof(p) && (i < 99) )
{
fread(&buf[i++],1,1,p);
}
buf[i] = 0;
printf("%s",buf);
pclose(p);
return 0;
}
else
{
return -1;
}
}

Related

Why is this code correct while it should clearly run into an infinite loop?

I have been having a problem with this code for a while. The placement of recursive call of the function does not seem right.
i tried running the code and yes it does run into a infinite loop.
// I DEFINE HEAP STRUCTURE AS :
struct heap_array
{
int *array; // heap implementation using arrays(note : heap is atype of a tree).
int capacity; // how much the heap can hold.
int size; //how much size is currently occupied.
void MaxHeapify(struct heap_array *h,int loc) // note : loc is the location of element to be PERCOLATED DOWN.
{
int left,right,max_loc=loc;
left=left_loc_child(h,loc);
right=right_loc_child(h,loc);
if(left !=-1 && h->array[left]>h->array[loc])
{
max_loc=left;
}
if(right!=-1 && h->array[right]>h->array[max_loc])
{
max_loc=right;
}
if(max_loc!=loc) //i.e. if changes were made:
{
//swap the element at max_loc and loc
int temp=h->array[max_loc];
h->array[max_loc]=h->array[loc];
h->array[loc]=temp;
}
MaxHeapify(h,max_loc); // <-- i feel that this recursive call is misplaced. I have seen the exact same code in almost all the online videos and some books i referred to. ALSO I THINK THAT THE CALL SHOULD BE MADE WITHIN THE SCOPE OF condition if(max_loc!=loc).
//if no changes made, end the func right there.
}
In your current implementation, it looks like you don't have a base case for recursion to stop.
Remember that you need a base case in a recursive function (in this case, your MaxHeapify function), and it doesn't look like there is one.
Here is an example of MaxHeap which may be resourceful to look at
// A recursive function to max heapify the given
// subtree. This function assumes that the left and
// right subtrees are already heapified, we only need
// to fix the root.
private void maxHeapify(int pos)
{
if (isLeaf(pos))
return;
if (Heap[pos] < Heap[leftChild(pos)] ||
Heap[pos] < Heap[rightChild(pos)]) {
if (Heap[leftChild(pos)] > Heap[rightChild(pos)]) {
swap(pos, leftChild(pos));
maxHeapify(leftChild(pos));
}
else {
swap(pos, rightChild(pos));
maxHeapify(rightChild(pos));
}
}
}
Here, you can see the basecase of:
if (isLeaf(pos))
return;
You need to add a base case to your recursive function.

protothread jump a thread to the beginning in the external main function

I have a protothread set up and blocking ...
static int mythread(struct pt *pt){
static int k;
PT_BEGIN(pt)
while(1){
PT_WAIT_UNTIL(pt, eventA == 1); // blocked at lineA
for(k=0;k<100;k++){
//do something
PT_YIELD(pt); //blocked at lineB
}
PT_WAIT_UNTIL(pt, eventB == 1); //block at lineC
}
PT_END(pt)
}
After a while, mythread can be blocked at "lineA", "lineB", or "lineC".
How could an external function, like main() reset mythread to be blocked at the beginning "lineA" again.
By running the macro PT_RESTART(&pt_mythread)? The compiler doesn't like it. Because my main() function isn't inside PT_BEGIN, PT_END block, so the return inside that macro is bad, bad.
Or running PT_INIT(&pt_mythread) again? Any suggestions?
Yes, calling PT_INIT from outside the protothread will restart it. If you look at the source for PT_RESTART:
#define PT_RESTART(pt) \
do { \
PT_INIT(pt); \
return PT_WAITING; \
} while(0)
This is exactly what it does, but then also returns (like a yield) out of the thread. As you say it's designed to be called from inside the protothread.
The protothread struct is basically just a number representing where it was in the thread:
struct pt {
lc_t lc; // where lc_t is an unsigned short;
};
So the only thing we need to do is reset that number to zero, which is exactly what PT_INIT does.

How does fork and pipe work in a unix process system?

I'm trying to study for an exam and I'm just not able to figure out a simple fork program.
I have this piece of code and have to add code to it In order for the parent process to send through a PIPE the value n to the child. The child should double the value, not print anything and return it to the parent.
Then the parent should print it on the screen.
int main() {
int n=1;
if(fork() == 0) {
}
printf(“%d\n”, n);
return 1;
}
I don't really know how PIPEs work and how to use them. Can anyone help me?
pid_t cp;
int fi[2],st;
int n;
if(pipe(fi)==-1) {perror("pipe error");exit(0);}
if((cp=fork())==-1) {perror("fork"); exit(0);}
else if(cp==0)
{
sleep(2);
close(fi[1]);
read(fi[0],&n,2);
n*=2;
close(fi[0]);
exit(n);
}
else
{
close(fi[0]);
write(fi[1],n,2);
close(fi[1]);
waitpid(cp,&st,0);
printf("%d",st);
exit(0);
}}
The working of pipes is very simple. A PIPE contains two ends, 1 for reading and another for writing. You have to close the appropriate end while reading or writing. After that you use it as a regular file with read() and write() functions. Forgive me for my formatting, I'm typing on a mobile.

Error passing pointer in QT

In QT have the following code that starts a thread to send out commands. The thread takes a char * and int as arguments. In the "run" I use the pointer that is given by the constuctor. The code is:
MyThread::MyThread(char * payld, int payld_size)
{
payload_size = payld_size;
payload_p = payld;
}
void MyThread::run()
{
while(...)
{
sendCommand(payload_p, payload_size);
}
}
Unfortunately this doesn´t work and my application crashes when I try to use thread.start(). But when I change it to:
MyThread::MyThread(char * payld, int payld_size)
{
payload_size = payld_size;
payload_p = payld;
for(int i=0; i<payload_size; i++)
{
payload[i] = payld[i];
}
}
void MyThread::run()
{
while(...)
{
sendCommand(payload, payload_size);
}
}
The code does run and only crashes sometimes (looks pretty random to me). Can anybody Explain me why version one doesnt work and version two does? And any ideas on why the second code sometimes crashes? Could it be because the size of payload is not predefined (in the header file I defined it as
char payload[];
When I define it as:
char payload[10];
it seems to work better, but it is annoying to test since the crashes are pretty random.
instead of fiddling with char*, I would switch to QString (since you're using Qt). It takes a bit of learning, but it's almost mandatory to get code working smoothly in this framework. Then declare
QString payload;
and depending on sendCommand implementation, use one of the member functions QString to get the char*, like payload.toLatin1()

Need help in IPC through Pipes

I am Working On a lab.
A father process will create two son processes A and B.
Son A will send some string to son B through pipe.son B will Invert the String case of the String Got from Son A and will send back the Inverted string to son A.after receiving the inverted string son A will print it to the screen.
here is the code.
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <ctype.h>
void process_A(int input_pipe[], int output_pipe[])
{
int c;
char ch;
int rc;
close(input_pipe[1]);
close(output_pipe[0]);
while ((c = getchar()) > 0) {
ch = (char)c;
rc = write(output_pipe[1], &ch, 1);
if (rc == -1) {
perror("A_TO_B: write");
close(input_pipe[0]);
close(output_pipe[1]);
exit(1);
}
rc = read(input_pipe[0], &ch, 1);
c = (int)ch;
if (rc <= 0) {
perror("A_TO_B: read");
close(input_pipe[0]);
close(output_pipe[1]);
exit(1);
}
putchar(c);
}
close(input_pipe[0]);
close(output_pipe[1]);
exit(0);
}
void process_B(int input_pipe[], int output_pipe[])
{
int c;
char ch;
int rc;
close(input_pipe[1]);
close(output_pipe[0]);
while (read(input_pipe[0], &ch, 1) > 0) {
c = (int)ch;
if (isascii(c) && isupper(c))
c = tolower(c);
else if (isascii(c) && islower(c))
c = toupper(c);
ch = (char)c;
rc = write(output_pipe[1], &ch, 1);
if (rc == -1) {
perror("B_TO_A: write");
close(input_pipe[0]);
close(output_pipe[1]);
exit(1);
}
}
close(input_pipe[0]);
close(output_pipe[1]);
exit(0);
}
int main(int argc, char* argv[])
{
/* 2 arrays to contain file descriptors, for two pipes. */
int A_TO_B[2];
int B_TO_A[2];
int pid;
int rc,i,State;
/* first, create one pipe. */
rc = pipe(A_TO_B);
if (rc == -1) {
perror("main: pipe A_TO_B");
exit(1);
}
/* create another pipe. */
rc = pipe(B_TO_A);
if (rc == -1) {
perror("main: pipe B_TO_A");
exit(1);
}
for(i=0;i<2;i++)
{
if((pid=fork()) <0){perror("fork failed\n");};
if((i==0) && (pid ==0))
{
process_A(A_TO_B, B_TO_A);
}
else if((i==1)&&(pid==0))
{
process_B(B_TO_A, A_TO_B);
}
else if(pid>0)
{
wait( &State );
}
}
return 0;
}
the problem is When i run the program the Son B gets Block.
I need u guys help.
Thanks in advance.
OK, diagram:
initially: parent process: has
B_TO_A[0] and [1] open,
has A_TO_B[0] and [1] open
fork (makes copy)
parent: child (pid==0):
B_TO_A both open, A_TO_B both open call process_A: close unwanted pipe ends, loop
call wait(), wait for one child loop reads stdin, writes one pipe, reads other pipe
if we ever get here:
fork (makes copy)
parent: child (pid==0):
B_TO_A both open, A_TO_B both open call process_B: close unwanted pipe ends, loop
parent: both ends of both pipes open
call wait(), wait for one child loop reads one pipe, writes other pipe
First, you will usually not get to "if we ever get here" because the child running process_A() runs in a loop until either EOF on stdin (if that occurs first) or one of the pipe read/write calls fails (e.g., due to EOF on input_pipe[0]). Since the parent is still waiting in a wait() call, and has both ends of both pipes open, there's no EOF on the pipe (EOF on a pipe occurs after you read all the data written by all writers, and all dups of the write end have been closed). So the only way to get there is to hit EOF on stdin, so that the while loop does not run.
Second, if you do get around to forking again and doing process_B(), that child will also wait forever, because one write end of the pipe it's reading from is still open... in the parent! The parent won't close it, because the parent will be waiting forever in wait.
In general, what you need to do here is:
create two pipes (like you do now)
fork once, and run process_A() in the child
fork again (in the parent), and run process_B() in the (new) child
close both ends of both pipes (in the parent)
wait for both children now, after both have gotten started
The error handling gets a bit messy since you have to do something (such as kill() the first child) if you can't start the second child. So you need to know how far along you have gotten. You can still loop to fork twice but you can't wait inside the loop, and with just two trips around the loop, each of which do rather different steps, you might as well just write it all out without a loop.

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