为什么的sizeof（int）的==的sizeof（长）？ [英] Why is the sizeof(int) == sizeof(long)?

问题描述

在下面列出的程序下，sizeof（int）和sizeof的（长）是我的机器（包括等于4个字节（或32位））在平等的。长，据我所知，是8个字节。它是否正确？我有一个64位计算机

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆limits.h中GT;诠释主要（无效）{
    的printf（的sizeof（短）=％d个\\ N的sizeof（短））;
    的printf（的sizeof（int）的=％d个\\ N的sizeof（INT））;
    的printf（的sizeof（长）=％d个\\ N的sizeof（长））;
    的printf（的sizeof（浮动）=％d个\\ N的sizeof（浮动））;
    的printf（的sizeof（双）=％d个\\ N的sizeof（双））;
    的printf（的sizeof（长双）=％d个\\ N的sizeof（长双））;
    返回0;
}
 

解决方案

  

有一个很长，据我所知，是8个字节。这是正确的？

没有，这是不正确的。在C和C ++规范仅指出长必须大于或等于32位。 INT 可以更小，但在许多平台上，C和C ++，长和 INT 都是32位。

这是一个很好的理由 preFER固定宽度整数类型如的int64_t 如果他们提供给你，你正在使用C99或为您提供一个equivelent类型的框架。

In the program listed below, the sizeof(int) and sizeof(long) are equal on my machine (both equal 4 bytes (or 32 bits)). A long, as far as I know, is 8 bytes. Is this correct? I have a 64-bit machine

#include <stdio.h>
#include <limits.h>

int main(void){
    printf("sizeof(short) = %d\n", sizeof(short));
    printf("sizeof(int) = %d\n", sizeof(int));
    printf("sizeof(long) = %d\n", sizeof(long));
    printf("sizeof(float) = %d\n", sizeof(float));
    printf("sizeof(double) = %d\n", sizeof(double));
    printf("sizeof(long double) = %d\n", sizeof(long double));
    return 0;
}

解决方案

A long, as far as I know, is 8 bytes. Is this correct?

No, this is not correct. The C and C++ specifications only state that long must be greater than or equal to 32 bits. int can be smaller, but on many platforms, in C and C++, long and int are both 32 bits.

This is a very good reason to prefer fixed width integer types such as int64_t if they're available to you and you're using C99 or a framework which provides you an equivelent type.

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