Difference between two struct initializations

Question

What is the difference between the following two initializations for a Struct?

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

And:

Car dodge = (Car) {
    .name = "Ram",
    .price = 1000
};

From Compiler Explorer, it looks like the two produce the same code:

What does the (StructName) do when preceding the struct? It seems its necessary when doing complex initializations such as:

CarPtr mazda = & (Car) {
    .name = "Mazda",
    .price = 20000
};

Also related, to the two answers from Possible to initialize/assign a struct pointer?.

See this Q/A about compound literals. Perhaps not the best post, but there are many others on compound literals in C. — Adrian Mole
– Adrian Mole, Commented Jan 6, 2021 at 22:23
It's not necessary when initializing a struct variable. You need it when initializing a pointer, because the pointer needs an object to point to, and the compound literal creates an object. — Barmar
– Barmar, Commented Jan 6, 2021 at 22:25
@Barmar wouldn't the object be {...} ? Why would the (ObjType) be needed to specify that? — David542
– David542, Commented Jan 6, 2021 at 22:28
You can't create an object with just {...}. That's an initializer list, so it can only be used when initializing a struct. Compound literals were invented to allow anonymous objects to be created. — Barmar
– Barmar, Commented Jan 6, 2021 at 22:29

Vlad from Moscow · Accepted Answer · 2021-01-06 22:40:12Z

3

In this declaration

Car dodge = (Car) {
    .name = "Ram",
    .price = 1000
};

there are created two objects of the type Car. The first one is the unnamed compound literal

(Car) {
    .name = "Ram",
    .price = 1000
}

that is used to initialize another named object dodge.

From C Standard (6.5.2.5 Compound literals)

3 A postfix expression that consists of a parenthesized type name followed by a braceenclosed list of initializers is a compound literal. It provides an unnamed object whose value is given by the initializer list.

In fact it is similar to the following declarations

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

Car dodge = ford;

The difference is that in the last example we created one more named object.

From the C Standard (6.7.9 Initialization)

13 The initializer for a structure or union object that has automatic storage duration shall be either an initializer list as described below, or a single expression that has compatible structure or union type. In the latter case, the initial value of the object, including unnamed members, is that of the expression.

edited Jan 6, 2021 at 22:40

answered Jan 6, 2021 at 22:25

Vlad from Moscow

313k27 gold badges204 silver badges358 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

David542 Over a year ago

would both initializations take up the same amount of memory in storage, or does creating the second object double it or?

Vlad from Moscow Over a year ago

@David542 The compound literal will be alive in the scope where it is defined. That is as I wrote you created two objects.

Collectives™ on Stack Overflow

Difference between two struct initializations

1 Answer 1

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related