Question

I am trying to do some simple updating of a string in a function, I got the following sample to work:

void change(char* buffer) {
  buffer[0] = 'b';
}

void main() {
  char buffer[20] = "abc   def   ghi   j\0";
  printf("before: .%s., %p\n", buffer, buffer);
  change(buffer);
  printf("after:  .%s.\n", buffer);
}

But if I use char* for buffer instead of char[], I get an error in the function. So the following example doesn’t work:

void compact(char* buffer) {
  buffer[0] = 'b';
}

void main() {
  char* buffer="abc   def ghi\0";
  printf("before: .%s., %p\n", buffer, buffer);
  change(buffer);
  printf("after:  .%s.\n", buffer);
}

Any suggestions on what I am doing wrong? Thanks.

Don

Answer 1

If you use char*, your variable points to a string literal and invokes undefined behaviour.

You can’t modify a string literal. When you declare a pointer like this

char* buffer="abc   def ghi\0";

buffer will point to read-only memory (probably, implementation defined). Regardless, you can’t change its contents. This syntax was kept for compatibility with C, but, IMO, this shouldn’t compile. Whenever you see something like that (char* pointing to a C-string), immediately translate it, inside your head, to

const char* buffer="abc   def ghi\0";

Put on your C++ glasses. If you don’t have them, get a pair (you’ll see all sorts of stuff, like this, returning locals by reference, forgetting to return, and many more).

The other case:

char buffer[] = "abc   def ghi\0";

is fine because you actually create an array of the required size and populate it with that content. This is no longer a pointer to a string literal.