C++ 冲突声明

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/11890083/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-08-27 15:40:32  来源:igfitidea点击:

Conflicting declaration

c++typedefconflictsize-t

提问by user1496542

I have a typedef defined in my code as

我在我的代码中定义了一个 typedef

typdef unsigned int size_t;

it is conflicting with stddef's

它与 stddef 相冲突

typedef __SIZE_TYPE__ size_t;

I'm unsure how to get around this but would still like to keep size_t in my code.

我不确定如何解决这个问题,但仍想在我的代码中保留 size_t。

回答by Luchian Grigore

TwoThree options:

三个选项:

1) Pick a different name, I think you already got that.

1)选择一个不同的名字,我想你已经知道了。

2) Use a namespace:

2)使用namespace

namespace X
{
   typedef long size_t;
}

and the type as

和类型为

X::size_t x;

3) Ugly, guaranteed to get you fired, and me downvoted:

3)丑陋,保证让你被解雇,我投反对票:

typedef unsigned int my_size_t;
#define size_t my_size_t

回答by jwismar

It is probably a bad idea to try to redefine a type that's in one of the standard headers. What are you trying to accomplish? Why don't you want to use the standard size_t definition?

尝试重新定义标准头文件之一中的类型可能是一个坏主意。你想达到什么目的?为什么不想使用标准的 size_t 定义?