problem with shared memory

Question

Tried to create shared memory block, and what I got is strange behaviour.

#include <sys/shm.h>
#include "stdio.h"
#include <sys/ipc.h>

int main() {
printf("starting\n");

int mid = -1;
mid = shmget((key_t)1234, 4096, IPC_CREAT|0666);
if(mid == -1) {
    printf("cant get mid\n");
    return 1;
} else {
    printf("got mid");
}

int* maddr = 0;
maddr = shmat(mid, NULL ,0);
if(maddr == (int*)-1) {
    printf("cant attach memory\n");
    return 1;
} else {
    printf("got maddr");
}

while(1) {
    int cval = __sync_add_and_fetch(maddr, 1);
    if(cval % 2) { // odd values
            printf("messager 1");
            sleep(1000);
    }
}
}

If I try to execute that code, it prints starting and hangs, nothing more happens, but some why it accepts input from stdin, so I can print just like if scanf is running

caf · Accepted Answer · 2011-05-27 11:32:08Z

4

stdout is line-buffered by default, which means that it is not flushed until a newline is printed. This means that you need to put \n at the end of your "got mid", "got maddr" and "messager 1" strings, or fflush(); after those printf()s.

By the way, SYSV shared memory is outdated. The POSIX mechanisms are significantly better designed - see shm_open() and related calls.

answered May 27, 2011 at 11:32

caf

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pickypg · Accepted Answer · 2011-05-27 11:32:14Z

3

Try adding a new line (\n) at the end of all of your printf statements. I assume it's failing before printing/flushing the buffer.

You can enter things in because the terminal is not blocking keystrokes, even though you haven't paused anywhere to read it. I regularly type in the "next thing" while my command-line/terminal is busy doing something else so that it just picks up when it's done. The stdin buffer can still accept the input. It just doesn't use it yet.

answered May 27, 2011 at 11:32

pickypg

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