char * ptr上的printf％s是否尊重它？

Question

I'm not sure if there exists a similar question, but

char *ptr = "xyz";
printf("%p\n", ptr);

prints an address, but (note the format specifiers)

char *ptr = "xyz";
printf("%s\n", ptr);

prints the string?

So far what I've learnt is that printing ptr would give the address, and printing *ptr would give the value. I'm quite confused.

Answer 1

First, check about the format specifier(s).

For %s , with fprintf() family, from C11 , chapter §7.21.6.1 ( emphasis mine )

s

If no l length modifier is present, the argument shall be a pointer to the initial element of an array of character type. ²⁸⁰⁾ Characters from the array are written up to (but not including) the terminating null character. [...]

and, for %p ,

p

The argument shall be a pointer to void . The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

So, basically, yes, when you attempt to print the content, you have to dereference the pointer (anyway).

Answer 2

When you print with %s in C. You are printing an array of characters up to the first '\\0' by definition of what a string is in C. So you actually use the adress of the first character and the printf will deference it away.

This means that

char *ptr = "xyz";
printf("%s\n", ptr);

will print from the address of the ptr until the first cell from ptr where ptr[x] == '\\0'

Answer 3

When using "%s" then printf will in essence loop over the string to print it character by character, until the string null-terminator character is found.

For that it needs to use array indexing, which indeed dereference the pointer.

With "%s" then printf uses the value if where the pointer is pointing at. With "%p" the value of the pointer itself is used.