是什么导致这个strcpy segfault？

Question

Here is my code and it faults here strcpy(pSrcString,"muppet"); Well infact it does whenever I use strcpy.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{

char *pSrcString = NULL;
char *pDstString = NULL;

/* muppet == 6, so +1 for '\0' */
if ((pSrcString = malloc(7) == NULL))
{
    printf("pSrcString malloc error\n");
    return EXIT_FAILURE;
}

if ((pDstString = malloc(7) == NULL))
{
    printf("pDstString malloc error\n");
    return EXIT_FAILURE;
}

strcpy(pSrcString,"muppet");

strcpy(pDstString,pSrcString);

printf("pSrcString= %s\n",pSrcString);
printf("pDstString = %s\n",pDstString);
free(pSrcString);
free(pDstString);

return EXIT_SUCCESS;
}

Answer 1

You have misplaced your parentheses in (pSrcString = malloc(7) == NULL) . In that way, you are first checking the result of malloc(7) against NULL (which turns out to be false or 0 ), and then assigns that to pSrcString . Basically:

pSrcString = 0;

Well of course that is not going to give you a valid memory to let strcpy write things to. Try this instead:

(pSrcString = malloc(7)) == NULL

Likewise for pDstString .

Aside, if you just want to have a copy of the string, you can use the strdup function. That allocates memory for you and takes care of calculating the length itself:

pSrcString = strdup("muppet");