8

I am trying to initalize a structure using braces, but i am really trying to initalize the structure that is pointed to by a pointer returned from a malloc call.

typedef struct foo{
    int x;
    int y;
} foo;

foo bar = {5,6};

I understand how to do that, but i need to do it in this context.

foo * bar = malloc(sizeof(foo));
*bar = {3,4};
Nick
  • 1,417
  • 1
  • 14
  • 21
AndrewGrant
  • 786
  • 7
  • 17

2 Answers2

8

(This was answered in comments, so making it a CW).

You need to cast the right-hand side of the assignment, like so:

*bar = (foo) {3,4};

As pointed out by @cremno in the comment, this isn't a cast but rather an assignment of a compound literal

The relevant section of the C99 standard is: 6.5.2.5 Compound literals which 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

jpw
  • 44,361
  • 6
  • 66
  • 86
  • 6
    It may look like a cast operator but it isn't one. The whole thing is a compound literal. – cremno Jul 15 '15 at 23:37
  • @cremno So it would seem. Good point. I've updated the answer with a reference to the standard. – jpw Jul 15 '15 at 23:41
  • As a notable bonus: any struct members you forgot to include in a compound literal initializer get zero-initialized by default (just like if they were static objects). See: https://stackoverflow.com/a/24936301/9959012 – 6equj5 Nov 25 '22 at 19:11
1

bar is a pointer that holds reference to the malloced foo struct

Use bar->x=3;bar->y=4

Shreyas Chavan
  • 1,079
  • 1
  • 7
  • 17