Question

Why the following code works:

char *p;

p="hello";

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

While this one doesn’t:

char *p;

strcpy(p,"hello");

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

I know that adding p=malloc(4); on the second example would make the code work, but that is exactly my question. Why malloc is needed in the second example but not in the first?

I looked for similar questions on SO but none answer this exactly.

Answer 1

p is a pointer. You need to make it point to something. In the first case,

 p = "hello";

makes p point to that string literal which is located somewhere in your program’s memory at runtime.

In your second case, you didn’t make p point to anything, so doing anything that looks at where p points to is invalid.

p = malloc(some_size);

makes p point to a piece of (uninitialized) memory that can hold some_size chars. If you reserved enough, you can then do things like strcpy(p, "hello") because p does point to a valid memory area, so copying into the memory pointed-to by p is ok. Note that some_size must be at least as big as what you want to copy into it, including the '\0' string terminator.

Note that doing:

p = "hello";
strcpy(p, "bye");

would be invalid because "hello" can be stored as in a read-only memory, so you can’t overwrite it.