c - int 是 4 个字节，但仍然可以存储在 char 中，为什么没有溢出

Question

检查这个程序

#include<stdio.h>

int main (){

char c='a';
printf("%d %d", sizeof(c),sizeof('a'));
}

输出是1 4
我知道当我们写一个语句时 char c='a';

那怎么会在 1 个字节的空间(char c)中存储一些 4 个字节的东西(ASCII 码)，为什么没有溢出等等。

Answer 1

首先，根据 ANSI/IEC 9899:1999(E) §6.4.4.4:

 10. An integer character constant has type int. The value of an integer character constant containing a single character that maps to a single-byte execution character is the numerical value of the representation of the mapped character interpreted as an integer. [...]

§6.5.3.4:

 2. The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. [...]

 3. When applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1. [...]

如您所见，由于字符常量的类型是 int，对于 sizeof('a') 我们得到 sizeof(int)，在你的平台上是 4。然而，对于 sizeof(c)，我们得到一个 char 的大小，它被定义为 1。

那么为什么我们可以将'a'赋值给char呢？

§6.5.16.1:

 2. In simple assignment (=), the value of the right operand is converted to the type of the assignment expression and replaces the value stored in the object designated by the left operand.

因此，'a' 的 int 被隐式转换为 char。那里也有一个示例，明确显示 int 可以隐式转换为 char。