c - 我是 c 编程初学者，需要 sizeof() 字符串常量方面的帮助吗？

Question

/**** Program to find the sizeof string literal ****/

#include<stdio.h>

int main(void)
{
printf("%d\n",sizeof("a")); 
/***The string literal here consist of a character and null character,
    so in memory the ascii values of both the characters (0 and 97) will be 
    stored respectively  which are found to be in integer datatype each 
    occupying 4 bytes. why is the compiler returning me 2 bytes instead of 8 bytes?***/

return 0;
}

输出:

2

Answer 1

字符串文字"a" 的类型为char[2]。你可以想象它像下面的定义

char string_literal[] = { 'a', '\0' };

sizeof( char[2] ) 等于 2 因为(C 标准，6.5.3.4 sizeof 和alignof 运算符)

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

C 中的字符常量确实具有 int 类型。例如 sizeof( 'a' ) 等于 sizeof( int ) 并且通常等于 4。

但是当 char 类型的对象由这样的字符常量初始化时

char c = 'a';

应用隐式缩小转换。