与printf一起使用时，C中的char *和int *之间的区别

Question

This could be a very basic question, but I am not able to understand it. I need a clear understanding, hence am posting here.

Consider the code:

char *c = "hello";
int   a = 10;
int  *b = &a;
printf("%s\t%d\n", c, *b);

Why do I have to pass *b to get value to be printed but in case of strings if I just give c printf still prints "hello"?

Answer 1

That's just what the specifiers mean:

• %s wants a pointer to a char, so you don't have to dereference it, printf will
• %d wants a true integer so you do have to dereference it

It's just what printf is and always was.

Answer 2

Because that's how printf is defined.

But it makes sense; consider the following code:

printf("%d\n", 42);

What would the equivalent code be if printf took integers via pointer?

Strings are a special case; a string in C is a sequence of characters in memory, accessed via a pointer to its first element. So you need to give printf that pointer, so that it can read the whole string.