Question

I have this code:

#define ABC "abc"

void main()
{
char *s = malloc(sizeof(char)*3);
printf("%p ", s);
s = ABC;
printf("%p ", s);
free(s);
}

This is the output:

0x8927008 0x8048574 Segmentation fault (core dumped)

As you can see, the address of string s changes after assignment (I think this is why free() gives segfault).
Can anyone explain me why and how this happens?
Thank you!

Answer 1

The line

s = ABC;

changes s to point to a different string which may well be in read-only memory. Attempting to free such memory results in undefined behaviour. A crash is likely.

I think you wanted

strcpy(s, ABC);

instead. This would copy the char array “abc” into s. Note that this will cause a further bug – s is too short and doesn’t have space for the nul terminator at the end of ABC. Change you allocation to 4 bytes to fix this

char *s = malloc(4);

or use

char *s = malloc(sizeof(ABC));

if ABC is the max length you want to store.