当我们将值重新分配给 char 指针时，内存会发生什么？

Question

I wonder what happens inside the memory when we do something like this

  char *s;
  s = "Text";
  s = "Another Text";

If I'm getting it right, by assigning string to char pointer memory is dynamically allocated. So according to my understanding assignment expression

  s = "Text";

equals to

  s = (char *) malloc(5); // "Text" + '\0'
  strcpy(s, "Text");

Well, this way we can easily free memory by using

  free(s);

But... After reassigning same pointer to another value, it allocates new memory segment to store that value.

  s = "Text";
  printf("\n(%p) s = \"%s\"", s, s);

  s = "Another Text";
  printf("\n(%p) s = \"%s\"", s, s);

Output:

  (0x400614) s = "Text"
  (0x400628) s = "Another Text"

That means that address of old value is not accessible to us any longer and we can't free that any more. Another call to free(s); will probably deallocate only last memory segment used by that pointer.

My question is: If we reassign same char pointer over and over again, does it consume more and more program memory during run-time or that garbage somehow gets automatically freed?

I hope that was enough to demonstrate my problem, couldn't think better example. If something's not clear enough please ask for additional clarification.

Answer 1

Your understanding is wrong. It is just the assignment and it does not allocate any memory. In your example you assign the pointer with the addresses of the string literals. String literals are created compile time and placed in the read only memory

You do now allocate any memory by assigning the pointer

Answer 2

It's not equal to doing a malloc. What's happening is that the string literal is stored in a read only part of memory. And it's not the assignment of a pointer that does the allocation. All string literals in a program are already allocated from start.

It might be worth mentioning that they are not strictly speaking stored in read only memory, but they might be and writing to a string literal is undefined behavior.

You cannot and should not call free on a string literal. Well, you can, but the program will likely crash.

Answer 3

• With no optimization, compiler will reserve two distinct memory space for string literals "text1" and "text2".

• If assignment lines are very consecutive as in your question and if nothing is done after the first assignment line —assuming compiling with optimization— compiler, most probably, will not allocate any space for the first string literal nor will produce any opcode for the first assignment line.