Initialize struct without an assignment?

Question

I could not find an answer to this on the Internet, so here is my question: Can I define a struct instance without assigning it to a local or global variable in C? E.g.:

struct A {
  int b;
}

struct A foo() {
  return struct A { .b = 42 };
}

If this is not possible: why?

https://nickdesaulniers.github.io/blog/2013/07/25/designated-initialization-with-pointers-in-c/ — Engineer2021, Apr 25 '14 at 13:42
BTW, functions taking no arguments should have `(void)` as the parameters - otherwise a prototype isn't formed, and the compiler will let you call it with arguments (causing undefined behaviour) — M.M, Apr 25 '14 at 13:54
@Matt MCNabb Thanks for the hint, I was definitely not aware of that! — Bastian, Apr 25 '14 at 14:01
possible duplicate of [How to initialize a struct in ANSI C](http://stackoverflow.com/questions/330793/how-to-initialize-a-struct-in-ansi-c) — Deduplicator, Apr 25 '14 at 14:04

Shafik Yaghmour · Accepted Answer · 2014-04-25T13:50:35.250

Yes C99 provides compound literals for this (see it live):

return (struct A) {  42 } ;

which is covered in the draft C99 standard section 6.5.2.5 Compound literals and says:

A postfix expression that consists of a parenthesized type name followed by a brace enclosed list of initializers is a compound literal. It provides an unnamed object whose value is given by the initializer list.⁸⁴⁾

and:

The value of the compound literal is that of an unnamed object initialized by the initializer list. If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block.

and provides several examples including:

EXAMPLE 3 Initializers with designations can be combined with compound literals. Structure objects created using compound literals can be passed to functions without depending on member order:

drawline((struct point){.x=1, .y=1}, (struct point){.x=3, .y=4});

gcc also has a nice document on this in it's extension section since it supports this feature outside of C99 as well as clang.

score 2 · Answer 2 · answered Apr 25 '14 at 13:40

2

Yes, you can use compound literals in C99 and later.

return (struct A) { .b = 42 };

You can even point to them:

struct A *a = &(struct A) { .b = 42 };
a->b = 43;

These literals are "better" than string literals in that they are writable. The compiler may pool them if and only if you include const in the literal's type .

answered Apr 25 '14 at 13:40

M.M

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you can also do `return (struct A) { 42 };`. – Engineer2021 Apr 25 '14 at 13:43
1

Of course; although I figured he wanted to use designated initializers to avoid errors in case `A` had other members added. – M.M Apr 25 '14 at 13:44
Yep, in that case it is better to be explicit. – Engineer2021 Apr 25 '14 at 13:44

Deduplicator · Answer 3 · 2014-04-25T13:58:49.200

Yes, it is possible since C99. That is a compound literal.

Still, yours has the wrong syntax. Use:

(struct A){.b=42}

or

(struct A){42}

Though, go for constant literals if it does not matter:

(const struct A){.b=42}

All constant literals are subject to constant pooling (including string literals which have type char[] for historical reasons).
Constant compound literals and compound literals outside any function have static storage duration,
the rest have automatic storage duration (beware returning a pointer, also they must be initialised each time).

In summary, prefer constant literals where possible.

Initialize struct without an assignment?

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