为什么在没有类型转换的情况下复合文字赋值不起作用

我有一个关于 C 语言文字的问题。

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，这似乎不起作用？这是什么原因呢？

(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后缀表达式 初始值设定项列表是复合文字。