Question

#include <stdio.h>
int main()
{
    char str[11] = "HelloWorld";
    printf("%s\n",str);
    printf("%s\n",str+3);

    /* This Line here is the devil */
    printf("%s\n",str[2]); // %s needs an addr not a value.

    return 0;
}

Why does that line give a segmentation fault. Is it because %s in printf needs an address and not a value.
What is the actual reason ??

Answer 1

str[2] returns a char, not a pointer to a char. So, printf will try to start reading at address 0x6c. Right there, there is a good chance that 0x6c is an invalid address that will cause a segfault. However, if it isn’t invalid then printf will keep reading until it reaches a 0x00 character, which very well could enter into an invalid address range.

If you want to know precisely why it segfaults, you would need to follow along in a debugger, which might be interesting and educational.

If you wanted to fix the crashing line, you could change it to:

  printf("%s\n", &str[2]);

which I would consider to be better style than str+2.