Question

Why does the following happen:

char s[2] = "a";
strcpy(s,"b");
printf("%s",s);

–> executed without problem

char *s = "a";
strcpy(s,"b");
printf("%s",s);

–> segfault

Shouldn’t the second variation also allocate 2 bytes of memory for s and thus have enough memory to copy "b" there?

Answer 1

Shouldn’t the second variation also allocate 2 bytes of memory for s and thus have enough memory to copy “b” there?

No, char *s is pointing to a static memory address containing the string "a" (writing to that location results in the segfault you are experiencing) whereas char s[2]; itself provides the space required for the string.

If you want to manually allocate the space for your string you can use dynamic allocation:

char *s = strdup("a"); /* or malloc(sizeof(char)*2); */
strcpy(s,"b");
printf("%s",s); /* should work fine */

Don’t forget to free() your string afterwards.