在C函数中传递指针参数

Question

#include <stdio.h>

// this works
void print_stuff (void* buf) {
    printf ("passed arg as buf*: %s\n", buf);
}

/* This works */
void print_stuff_3 (char* buf) {
    printf ("passed arg as char*: %s\n", buf);
}

// this does not work
void print_stuff_2 (char** buf) {
    printf ("%s\n", *buf);
}

int main () {
    char s [] = "hi";

    printf ("s = %s\n", s); 

    // these work
    print_stuff (&s);
    print_stuff_3 (&s);

    // this results in a Segfault
    print_stuff_2(&s);
    return 0;
}

I am a bit confused about the way things are passed around in C. I feel like &s should be of type char** , but it behaves as if it is of type char* when passed to a function. Why does this behaviour happen?

In particular, print_stuff_2 segfaults, whereas I thought that print_stuff_3 would give an error.

EDIT: To clarify, I expected print_stuff(&s) and print_stuff_3(&s) to fail (while they succeed), while print_stuff_2(&s) fails, whereas I feel it should succeed.

Answer 1

You need to remember that strings are not fundamental types in C. They are arrays of characters. Therefore

char s [] = "hi";

makes s a char * (in terms of variable type), ie a pointer to the first character of a 3 character array ( h , i and NUL ).

So in order to pass a pointer to the string, you what to use your print_stuff_3 , as printf() 's %s argument takes exactly that (a pointer to the string, ie a pointer to the first character). Call this with print_stuff_3(s) .

print_stuff works because a pointer is a pointer. It will be translated to a void * pointer on calling print_stuff , then printf() 's %s will convert it back to a char * . Call this with print_stuff(s) .

print_stuff_2 doesn't work because you are taking the address of where s is stored. Had you written char *s = "hi"; that would work if you used print_stuff_2(&s) . You'd pass the address of the pointer, then dereference that (to get the value of the pointer, ie the pointer to the first character) in by using *buf . Except buf then would be a poor choice of name, as you would be passing a pointer to a pointer to characters.

The complication is as follows. As it is, you are doing &s which just returns s when you have

char s [] = "hi";

(see How come an array's address is equal to its value in C? ), but returns the address at which the pointer variable s is stored on the stack if you have:

char *s = "hi";

Taking the address of an array doesn't really make sense (so evaluates to the address of the first element). You need to use char *s = "hi"; if you want to take the address of the pointer.

Answer 2

In C, array names are decays to pointer to its first element when passed to a function in most cases. When passing s to the function print_stuff , s decays to pointer to h . No need to pass it with & . &s is of pointer to array ( char (*)[3] ) type, ie, it is giving the address of the entire array s .

In function call

print_stuff_3 (&s);  

your compiler should warn you

[Warning] passing argument 1 of 'print_stuff_3' from incompatible pointer type [enabled by default]   

I feel like &s should be of type char* *, but it behaves as if it is of type char* when passed to a function. Why does this behavior happen?

No. You thought wrong. &s is of type char (*)[3] .

Answer 3

void print_stuff (void* buf) & void print_stuff_3 (char* buf) In both functions, buf is of char * taking address as argument. Which should be print_stuff (s) & print_stuff_3 (s) respectively as s is the base address of char array s . So you shouldn't pass &s which is address of s .

As the below function buf is of type char ** , it will expect address of address like print_stuff_2(&s) provided your declaration is char *s = "hi" ,

void print_stuff_2 (char** buf) {
    printf ("%s\n", *buf);
}