c - 为什么没有类型转换就不能进行复合文字赋值

Question

我对 C 中的文字有疑问。

int a;
//a is an integer that is assigned an integer literal 414
a = 414;

float b;
//b is a float that is assigned a float literal of 3.14
b = 3.14;

struct point {
    int x,y;
};

struct point b;
//{5,6} is a compound literal that is assigned to a struture.
b = {5,6}; //doesn't work.

b = (struct point){5,6}; //works.

如果没有类型转换，这似乎行不通？这是什么原因？

Answer 1

(struct point){5,6}作为一个整体是一个复合文字。

C11 §6.5.2.5复合文字

由带括号的类型名称后跟大括号括起来的初始值设定项列表组成的后缀表达式是复合文字。