write to shared memory segmentation fault

Question

all I want to do is just write "hey" to my shared memory, but it gets thrown at that line. very simple code as follows:

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <sys/fcntl.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/sem.h>
#include <sys/shm.h>
#define SHM_SIZE 1024
#define FLAGS IPC_CREAT | 0644
int main(){
    key_t key;
    int shmid;
    if ((key = ftok("ex31.c", 'k')) == -1){ 
        exit(1);}        
    if ((shmid = shmget(key, SHM_SIZE, FLAGS)) == -1) {
            exit(1);}
    char* shmaddr;
    if( shmaddr=shmat(shmid,0,0) == (char*)-1){   //WRONG ARGUMENTS ??
            exit(0); }
    printf("opened shared memory\n");  //gets here
    strcpy(shmaddr, "hey");     //GETS THROWN HERE
    printf("after writing to memory\n"); //never get here

the error the debugger gives me is:

Program received signal SIGSEGV, Segmentation fault. 0x0000000000401966 in main (argc=1, argv=0x7fffffffe068) at ../ex31.c:449 449 strcpy(shmaddr, "hey"); //GETS THROWN HERE

Answer 1

The problem is operator precedence. In the line

shmaddr=shmat(shmid,0,0) == (char*)-1)

Then the unary - operator and the cast operator have the highest precedence, followed by == , followed by = which has the lowest precedence. So your code turns out to be equal to

shmaddr=(shmat(shmid,0,0) == (char*)-1))

which is nonsense.

Assignment inside conditions is very bad programming practice . Every half-decent compiler will give you a warning if you try it. Apart from operator precedence issues, it is easy to mix up = and ==. Also, the = operator introduces a side effect to the expression, so mixing it with other operators could cause undefined behavior. And perhaps most importantly, putting = inside conditions usually reduce code readability quite a lot.

Your code should be written as:

shmaddr = shmat(shmid,0,0);
if(chmaddr  == (char*)-1){

It is important to understand that you gain nothing by merging these 2 rows into 1. The generated machine code will be identical, apart from the lack of bugs in the above version.