为结构指针赋值的语法

Question

I have a quick question about the syntax and the code.

I just found out a way to declare the struct in C which is a bit different from what I've seen so far.

This is the sample code:

#include <stdio.h>
struct student {
    int age;
    int grade;
    char* name;
};
int main () {
    struct student s = { .name = "Max", .age = 15, .grade = 8 };
    return 0;
}

Assigning variables as .var_name works fine and you can assign it in any order you'd like. I liked this approach so I started experimenting and thus hit the wall. If I were to declare:

struct student *s = { ->name = "Max", ->age = 15, ->grade = 8 };

This won't work. Is there any syntax that could give the same results when using a pointer as the code above?

Thanks!

Answer 1

Generally speaking, you can't "assign values to a pointer". A pointer only stores a single value, which is the address of an already existing object.

If you have a pointer pointing to an existing struct variable, you can just initialize that:

struct student s = { .name = "Max", .age = 15, .grade = 8 };
struct student *p = &s;

If you want to use dynamic memory allocation, things get a bit tricker:

struct student *p = malloc(sizeof *p);
if (!p) {
    ...
}

malloc gives you uninitialized memory, and you cannot use initializer syntax with existing objects.

But you can use a trick involving compound literals (available since C99):

*p = (struct student){ .name = "Max", .age = 15, .grade = 8 };

Here we use a compound literal to create a new unnamed struct object whose contents we then copy into *p .

The same feature can be used to get rid of s in the first example:

struct student *p = &(struct student){ .name = "Max", .age = 15, .grade = 8 };

But in this version p still points to automatic memory (like a local variable).