Does making a char* point to another string literal leak memory?

Question

I Have some misunderstanding regarding this simple example:

char *s = "abc";
s = "def";

Will the assignment of the second string cause a memory leak, or it will be replaced properly? If the former is true, how do I replace the value of s the right way?

Answer 1

No. There's no memory leak.

You have simply changed s to point to a different string literal which is fine (with this: s = "def";).

Put simply, if you haven't allocated anything yourself using malloc/calloc/realloc or any function that returns a dynamically allocated memory (and documentation tells you to free(). For example, POSIX's getline() function does that), you don't need to free().

Answer 2

Both strings are defined statically so there's no memory leak. What would leak would be:

char *s = strdup("abc");  // dynamic memory allocation
s = "def";

Actually there's a small static memory waste (wouldn't call that a leak) since you're not able to use "abc" anymore.

But since it cannot be repeated even by calling the routine where your code is located, I definitely wouldn't call that a leak.

Now if you have char s[] = "abc"; (array, not pointer), and a "def" string of equal size, you could do:

strcpy(s,"def");

to replace the contents of the string (but don't do that on a statically assigned pointer like defined your code: undefined behaviour).