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} 整体是一个复合字面量。

C11 §6.5.2.5 复合文字

一个后缀表达式，由带括号的类型名称后跟一个大括号组成 初始化器列表是一个复合文字。