为什么“long int”与“int”具有相同的大小？这个修饰符是否有效？ [英] Why "long int" has same size as "int"? Does this modifier works at all?

问题描述

Ehm ..虽然像长 / 短这样的修饰符扩展/减少了创建变量时分配的内存量，但是... ...

Ehm.. I kind' of though this modifiers like long / short expands / reduces amount of memory allocated when variable are created, but...

#include <stdio.h>

#define test_int int
#define long_int long int
#define long_long_int long long int

void main()
{
    printf("%i\n", sizeof (test_int)); //output 4
    printf("%i\n", sizeof (long_int)); //output 4. Why? wasn't I modified it's size?
    printf("%i\n", sizeof (long_long_int)); //output 8
}

由于未知原因，它会打印的大小int 和 long int 一样。
我使用的是vc ++ 2010 express版。
很抱歉，很难在谷歌找到答案，它总是将长和 int 显示为单独的类型。

For unknown reasons, it prints the size of int and long int as same. I use vc++ 2010 express edition. Sorry, hard to find answer in google, it always shows long and int as separate types.

推荐答案

即使在64位系统上，MS选择使 long 32位的原因是现有的Windows API，历史原因使用 int 和 long 的混合物来表示类似的事情，并且期望这是32位价值（其中一些可以追溯到Windows是16位系统的时候）。因此，为了将旧代码转换为新的64位体系结构，他们选择将 long 保持为32位，以便混合 int <的应用程序/ code>和 long 在各个地方仍然会编译。

The reason that MS choose to makelong 32 bits even on a 64-bit system is that the existing Windows API, for historical reasons use a mixture of int and long for similar things, and the expectation is that this is s 32-bit value (some of this goes back to times when Windows was a 16-bit system). So to make the conversion of old code to the new 64-bit architecture, they choose to keep long at 32 bits, so that applications mixing int and long in various places would still compile.

C ++标准中没有任何内容这要求 long 应该大于 int （当然不是在大多数32位系统上）。所有标准都说大小为短< = int < = long - 并且 short 至少为16位，如果内存服务[不一定表示为应该至少为16位，我认为提到价值范围]。

There is nothing in the C++ standard that dictates that a long should be bigger than int (it certainly isn't on most 32-bit systems). All the standard says is that the size of short <= int <= long - and that short is at least 16 bits, if memory serves [not necessarily expressed as "should be at least 16 bits", I think it mentions the range of values].

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