在shell上的此c程序中为什么会出现分段错误错误

Question

why there is segmentation-fault error in the following code:

#include<stdio.h>

int main()
{
    char word[]="cs311cs312cs313";

    printf("%s %s %s %s",
           *word,
           *(word+2),
           *(word+4),
           *(word+8));

    return 0;
}

Is this error is due to compiler or there is some syntax error in my code ?

Answer 1

Modern compilers usually give warnings in such situations, for example:

warning: format specifies type 'char *' but the argument has type 'char' [-Wformat]

*(word+2) and others are pointer dereferencing , operation gets the actual value the pointer points to. This will give you chars 'c' , '3' and so.

Because printf has variable argument count - arguments after the format string will be put on stack raw, loosing types

%s makes printf take a few bytes from stack and expect it to be a pointer to the string. Try printf("Char: %p, str:%p", 'c', "c"); to see what the passed address looks like.

Addresses like 0x63 in the very beginning of memory are valid, but your program is forbidden from accessing these, because the segment is not assigned to it, so you get the segmentation-fault / access violation

So to fix the error - remove the unwanted dereferencing (the * s)