是否可以将一个简单的代码块转换为结构体？

Question

Consider this code :

#include <stdio.h>

int main()
{
  struct test {
    int i;
    int t;
  };

  struct test hello = (struct test) {
    .i = 0, 
    .t = 1
  };

  printf("%d, %d\n", hello.i, hello.t);

  return 0;
}

Output:

0, 1

my question is what is this line (struct test) {.i = 0, .t = 1 } doing?

is it casting a block of code to type struct test? is that even possible?

if not please tell me what is it doing, and thanks.

Answer 1

struct test hello = (struct test) { .i = 0, .t = 1 }; uses two features of C called compound literals and designated initializers .

The general form of a compound literal is: ( type-name ) { initializer-list } . (There may also be a comma after the list.) For example, these are compound literals:

(int) { 3 }
(int []) { 0, 1, 2 }
(union { float f; unsigned int u; }) = { 3.4 }

A compound literal is an object with no name.

In a plain initializer list, you simply list values for the items in the object being initialized. However, you can also use designated initializers. A designated initializer uses either the name of a structure member or the index of an array element to specify which part of the object is to be given the indicated value:

{ struct { int a, b, c; }) = { .b = 4, .c = 1, .a = 9 }
(int a[1024]) = { [473] = 1, [978] = -1 }

So, in struct test hello = (struct test) { .i = 0, .t = 1 }; , we are creating struct test with i initialized to 0, and t initialized to 1. Then this struct test is used to initialize another struct test named hello .

That particular use is pointless, since it creates a temporary object for the purpose of something that could have been done directly. Nominally, it creates a temporary struct test which is initialized and then copied into the struct test hello . Then the temporary struct test has no further use. The effect is the same as simply writing struct test hello = { .i = 0, .t = 1}; . That initializes hello without using a temporary object. However, a good compiler will optimize them to the same code.

Answer 2

In C (and many related programming languages), the curly braces often mean:

{ this is where the code starts
} this is where the code ends

They can also have several different meanings though. They don't only refer to code, but also to data structures:

struct point {
    int x;
    int y;
};

In this case they define where the data structure begins and ends.

struct point pythagoras = {
    .x = 3,
    .y = 4
};

And in this case they define where the initial contents of the variable begins and ends.

In conclusion, the { means begin, and the } means end. Nothing more.

The "code" that initializes your struct test is not really code, although it really looks like it. Here it is again:

struct test hello = (struct test) {
    .i = 0, 
    .t = 1
};

Usually, the (type) means a type cast. But, in this case the (struct test) is followed by an opening brace { , and this combination is called a compound literal . It's a literal for a variable of a compound data type.