Possible to initialize/assign a struct pointer?

Question

Let's say I have the following struct and two versions to initialize it:

#include <stdio.h>

typedef struct Car *CarPtr;
typedef struct Car {
        const char*     name;
        unsigned int    price;
} Car;

int main(void) {

    Car ford = {
        .name = "Ford F-150",
        .price = 25000
    };
    print_struct(&ford);

    // is this possible to do in a single assignment?
    CarPtr jeep = {
        .name = "Jeep",
        .price = 40000
    };
    print_struct(jeep);

}

Is the second version possible to do directly? Or do I need to do something along the lines of:

CarPtr jeep;
jeep->name = "Jeep";
jeep->price = 40000;

Or what's the proper way to initialize the CarPtr type directly?

Pointers need to be allocated before the pointee can be initialized/assigned — Eugene Sh.
– Eugene Sh., Commented Jan 6, 2021 at 22:00
You might want to read this stackoverflow.com/questions/750178/… — Ed Heal
– Ed Heal, Commented Jan 6, 2021 at 22:03
@Heal Sorry i had overlooked that. The comment had been edited. — Adnan Ahmed
– Adnan Ahmed, Commented Jan 6, 2021 at 22:11
Since its a pointer you will have to use malloc for CarPtr also you haven't assigned any Car type variable's address to CarPtr in the last example in code and using a constructor function such as Car_new that uses malloc is recommended for initialization. — Adnan Ahmed
– Adnan Ahmed, Commented Jan 6, 2021 at 22:13

Eric Postpischil · Accepted Answer · 2021-01-06 22:09:56Z

2

Pointers, other than null pointers, need to point to objects that exist (which can be memory that has been allocated but not yet filled in). You can define a pointer and initialize it to an object you create at the same time, using a compound literal:

CarPtr jeep = & (Car) { .name = "Jeep", .price = 40000 };

This is a special syntax, with the type name in parentheses followed by braces containing initializers just as in a definition. (It is not a cast.)

You can also pass a compound literal or its address directly to a function:

print_struct(& (Car) { .name = "Jeep", .price = 40000 });

However, this has limited uses, as, if this is done inside a function, the object is temporary; its lifetime lasts only as long as the current function execution lasts. You can use a compound literal outside a function, and then it lasts for all of program execution. You should have a general understanding and awareness of the properties of compound literals before using them.

edited Jan 6, 2021 at 22:09

answered Jan 6, 2021 at 22:08

Eric Postpischil

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1 Comment

David542 Over a year ago

I see -- thanks for that! I was trying to do &{ .name = "Jeep", .price = 40000} directly but I think I need to explicitly say that the struct is a Car for that to work.

Vlad from Moscow · Accepted Answer · 2021-01-06 22:14:28Z

CarPtr is a pointer (scalar object) that may be initialized by a braced list containing only a single expression.

So this initialization

CarPtr jeep = {
    .name = "Jeep",
    .price = 40000
};

is incorrect.

Something similar to what you are trying to achieve looks with using a compound literal the following way.

#include <stdio.h>

typedef struct Car *CarPtr;
typedef struct Car {
        const char*     name;
        unsigned int    price;
} Car;

int main( void )
{
    CarPtr jeep = &( Car ){ .name = "Jeep", .price = 40000 };
    
    printf( "%s - %u\n", jeep->name, jeep->price );
}

The program output is

Jeep - 40000

Pay attention to that you shall not free the pointed object by using the standard function free because the compound literal has automatic storage duration in this context.

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