为什么初始化函数返回null？

Question

I'm writing a program in C for the first time. I have a good bit of experience with C++, but C's reliance on pointers and the absence of new and delete are really throwing me off. I defined a simple data structure, and wrote a function that will initialize it (by taking a pointer). Here's my code:

//in Foo.h
#include <stdio.h>
#include <stdlib.h>

typedef struct Foo {
    struct Foo * members[25] ;
} Foo ;

void foo_init(Foo * f) ;
void foo_destroy(Foo * f) ;


//in Foo.c
void foo_init(Foo * f) {
    f = (Foo*)malloc(sizeof(Foo));

    for (size_t i = 0 ; i < 25 ; i++) {
        f->members[i] = NULL ;
    }
}

//define foo_destroy()

//in main.c
#include "Foo.h"

int main(int argc, const char * argv[])
{

    Foo * f ;
    foo_init(f) ;
    /* why is f still NULL here? */

    foo_destroy(f) ;

    /* ... */
    return 0;
}

When I tested my foo_init() function on f (pointer to a Foo struct), it was null after the function returned. The pointer f inside foo_init() is initialized just fine, however, so I don't think this is a problem with the init function itself. Shot in the dark, but could this be related to the way C handles passing by value/passing by reference (something I still don't entirely have a grasp on)? How can I correct this?

Answer 1

1) You go out of bounds in your for-loop, (as pointed out by @user3121023 change to 25 )

2) You are (m)allocating space for a local variable, use:

Foo *foo_init(void) {
    Foo *f = malloc(sizeof(Foo)); /* don't cast malloc */

    for (size_t i = 0 ; i < 25 ; i++) {
        f->members[i] = NULL ;
    }
    return f;
}

and in main :

f = foo_init();

Answer 2

void foo_init(Foo* f)

In C parameters are passed by value. Here you pass a parameter named f , of type Foo* . In the function you assign to f . But since the parameter is passed by value, you are assigning to the local copy, private to that function.

In order for the caller to see the newly allocated struct, you need an extra level of indirection:

void foo_init(Foo** f)
{
    *f = ...;
}

And at the call site:

Foo* f;
foo_init(&f);

Now, since your function is designed to send a new value to the caller, and your function currently has void return value, it would make more sense to return the new value to the caller. Like this:

Foo* foo_init(void)
{
    Foo* foo = ...;
    return foo;
}

You would call this like so:

Foo* f = foo_init();

Answer 3

In function void foo_init(Foo * f) you are passing a pointer. Now this pointer is passed as value in this function and modifications to this variable inside the function will not be reflected outside this function. In other words the variable f in function foo_init is a local variable.

so you should return the value of f by doing:

Foo* foo_init(Foo * f)
{
   f = (Foo*)malloc(sizeof(Foo));

   for (size_t i = 0 ; i < 25 ; i++) {
    f->members[i] = NULL ;
   }
   return f;
}

Apart from this note that I have changed the condition in for loop. You have allocated array of 25 objects. So your for loop should go from 0 to 24.

and in main you should

f = foo_init();