char和int的指针

Question

Hi I have a simple question

char *a="abc";
printf("%s\n",a);

int *b;
b=1;
printf("%d\n",b);

Why the first one works but the second one doesnot work? I think the first one should be

char *a="abc";
printf("%s\n",*a);

I think a stores the address of "abc". So why it shows abc when i print a? I think I should print *a to get the value of it.

Thanks

Ying

Answer 1

Why the first one works but the second one doesnot work?

Because in the first, you're not asking it to print a character, you're asking it to print a null-terminated array of characters as a string.

The confusion here is that you're thinking of strings as a "native type" in the same way as integers and characters. C doesn't work that way; a string is just a pointer to a bunch of characters ending with a null byte.

If you really want to think of strings as a native type (keeping in mind that they really aren't), think of it this way: the type of a string is char * , not char . So, printf("%s\\n", a); works because you're passing a char * to match with a format specifier indicating char * . To get the equivalent problems as with the second example, you'd need to pass a pointer to a string—that is, a char ** .

Alternatively, the equivalent of %d is not %s , but %c , which prints a single character. To use it, you do have to pass it a character. printf("%c\\n", a) will have the same problem as printf("%d\\n", b) .

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From your comment:

I think a stores the address of "abc". So why it shows abc when i print a? I think I should print *a to get the value of it.

This is where the loose thinking of strings as native objects falls down.

When you write this:

char *a = "abc";

What happens is that the compiler stores a array of four characters— 'a' , 'b' , 'c' , and '\\0' —somewhere, and a points at the first one, the a . As long as you remember that "abc" is really an array of four separate characters, it makes sense to think of a as a pointer to that thing (at least if you understand how arrays and pointer arithmetic work in C). But if you forget that, if you think a is pointing at a single address that holds a single object "abc" , it will confuse you.

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Quoting from the GNU printf man page (because the C standard isn't linkable):

d, i

The int argument is converted to signed decimal notation …

c

… the int argument is converted to an unsigned char, and the resulting character is written…

s

… The const char * argument is expected to be a pointer to an array of character type (pointer to a string). Characters from the array are written up to (but not including) a terminating null byte ('\\0') …

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One last thing:

You may be wondering how printf("%s", a) or strchr(a, 'b') or any other function can print or search the string when there is no such value as "the string".

They're using a convention: they take a pointer to a character, and print or search every character from there up to the first null. For example, you could write a print_string function like this:

void print_string(char *string) {
    while (*string) {
        printf("%c", *string);
        ++string;
    }
}

Or:

void print_string(char *string) {
    for (int i=0; string[i]; ++i) {
        printf("%c", string[i]);
    }
}

Either way, you're assuming the char * is a pointer to the start of an array of characters, instead of just to a single character, and printing each character in the array until you hit a null character. That's the "null-terminated string" convention that's baked into functions like printf , strstr , etc. throughout the standard library.

Answer 2

Strings aren't really "first-class citizens" in C. In reality, they're just implemented as null-terminated arrays of characters, with some syntactic sugar thrown in to make programmers' lives easier. That means when passing strings around, you'll mostly be doing it via char * variables - that is, pointers to null-terminated arrays of char .

This practice holds for calling printf , too. The %s format is matched with a char * parameter to print the string - just as you've seen. You could use *a , but you'd want to match that with a %c or an integer format to print just the single character pointed to.

Your second example is wrong for a couple of reasons. First, it's not legal to make the assignment b = 1 without an explicit cast in C - you'd need b = (int *)1 . Second, you're trying to print out a pointer, but you're using %d as a format string. That's wrong too - you should use %p like this: printf("%p\\n", (void *)b); .

What it really looks like you're trying to do in the second example is:

int b = 1;
int *p = &b;
printf("%d\n", *p);

That is, make a pointer to an integer, then dereference it and print it out.

Editorial note: You should get a good beginner C book (search around here and I'm sure you'll find suggestions) and work through it.