C: what is wrong with this char pointer assignment?

A newbie question:

I was practice assignment to a char pointer, but found there was nothing printed out. Here is the code:

#include <stdio.h>

int main (void)
{
    char * option_string = NULL;
    option_string = (char*)malloc(sizeof(5));
    memset(option_string, 0, sizeof(char) * 5);

    int j;
    for ( j = 0; j < 5; j++)
    {
        *option_string++ = 'a';
    }

    printf("print options_string: %s\n", option_string)开发者_运维知识库; //!nothing was printed out!
    free(option_string);
    return 0; 
}

Thanks in advance!

---

You increment the pointer option_string which in turn makes it point past the end of string.

Try malloc(6);

for ( j = 0; j < 5; j++)
{
    option_string[j] = 'a';
}

instead.

---

Your malloc(sizeof(5)) seems wrong. What is the size of 5? (Hint: It's not 5.)

---

Your problem is that within the loop, you write *option_string++. This means that once the loop is done, you're going to be pointing past the end of the string:

option_string at start
   |
   V
+----+----+----+----+----+
|    |    |    |    |    |
|    |    |    |    |    |
+----+----+----+----+----+
                             ^
                             |
                     option_string at end

Note that this reveals a second problem with your code: strings in C are null-terminated, but this string will eventually contain "aaaaa" and then… who knows? Garbage, most likely, but you can't tell. You need a six-length string. Fixing the first problem means using simple indexing instead: option_string[j] = 'a'. If you really want the *option_string++ method, you'll have to save and restore option_string (char * real_option_string = option_string; ... option_string = real_option_string;), but I wouldn't recommend it. Fixing both of these bugs, and a couple of style things, gives you:

#include <stdlib.h>
#include <stdio.h>

int main (void)
{
    char * option_string = calloc(6, sizeof(char));

    int j;
    for ( j = 0; j < 5; j++)
    {
        option_string[j] = 'a';
    }

    printf("print options_string: %s\n", option_string);
    free(option_string);
    return 0; 
}

The other thing I changed was your malloc usage. I feel like calloc is a better stylistic choice here; calloc(count, size) allocates count objects of size size, and zeros them. It's like malloc(count*size) plus a memset, but feels cleaner to me. You also shouldn't have a cast on malloc/calloc/etc., generally speaking (it can obscure useful warnings), and you need to allocate six slots, like I said (so you can have the null-terminator, which is the zero-valued character, so we don't need to set it explicitly). Combine that with the option_string[j] indexing mode, include the missing stdlib.h for calloc (you should have had it for malloc, too), and we're good to go!