Markus Markus - 1 month ago 9
C Question

Why do I get a segmentation fault when writing to a string initialized with "char *s" but not "char s[]"?

The following code receives seg fault on line 2:

char *str = "string";
str[0] = 'z';
printf("%s", str);


While this works perfectly well:

char str[] = "string";
str[0] = 'z';
printf("%s", str);


Tested with MSVC and GCC.

Answer

See the C FAQ, Question 1.32

Q: What is the difference between these initializations?
char a[] = "string literal";
char *p = "string literal";
My program crashes if I try to assign a new value to p[i].

A: A string literal (the formal term for a double-quoted string in C source) can be used in two slightly different ways:

  1. As the initializer for an array of char, as in the declaration of char a[] , it specifies the initial values of the characters in that array (and, if necessary, its size).
  2. Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element.

Some compilers have a switch controlling whether string literals are writable or not (for compiling old code), and some may have options to cause string literals to be formally treated as arrays of const char (for better error catching).