Dereference operator (*) differences between char* and int*

Question

Why is the operator (*) needed to access the value of an int* variable but not for char*?

char *char_ptr; 
int *int_ptr;
int mem_size = 50;

char_ptr = (char *) malloc(mem_size);
strcpy(char_ptr, "This is memory is located on the heap.");
printf("char_ptr (%p) --> '%s'\n", char_ptr, char_ptr);


int_ptr = (int *) malloc(12);
*int_ptr = 31337; 
printf("int_ptr (%p) --> %d\n", int_ptr, *int_ptr);

Output:

char_ptr (0x8742008) --> 'This is memory is located on the heap.' 
int_ptr (0x8742040) --> 31337

Answer 1

This is because of the way the printf format specifiers work: the %s format expects, for its corresponding argument, a pointer to a character (more precisely, the address of a nul-terminated string - which can be an array of char, an allocated buffer with at least one zero byte in it, or a string literal), so you can just give it the char_ptr variable as-is; on the other hand, the %d format expects an integer (not a pointer-to-integer), so you have to dereference the int_ptr variable using the * operator.

Note on Good Programming Style: As mentioned in the comments to your question, be sure to call free() at some point on every buffer allocated with malloc, or you will introduce memory leaks into your code. Also see: Do I cast the result of malloc?.