c++ - 重新定义 uint 类型 c++

Question

在 c++ 中，当我尝试编译以下代码时出现声明冲突错误:

#include <iostream>
using namespace std;

typedef uint_least64_t uint;

int main() {
    uint i = -1;
    cout << i << endl;
}

错误:

main.cpp:5:24: error: conflicting declaration ‘typedef uint_least64_t uint’
 typedef uint_least64_t uint;
                        ^~~~
In file included from /usr/include/stdlib.h:275:0,
                 from /usr/include/c++/6/cstdlib:75,
                 from /usr/include/c++/6/ext/string_conversions.h:41,
                 from /usr/include/c++/6/bits/basic_string.h:5417,
                 from /usr/include/c++/6/string:52,
                 from /usr/include/c++/6/bits/locale_classes.h:40,
                 from /usr/include/c++/6/bits/ios_base.h:41,
                 from /usr/include/c++/6/ios:42,
                 from /usr/include/c++/6/ostream:38,
                 from /usr/include/c++/6/iostream:39,
                 from main.cpp:1:
/usr/include/x86_64-linux-gnu/sys/types.h:152:22: note: previous declaration as ‘typedef unsigned int uint’
 typedef unsigned int uint;
                      ^~~~

我假设声明冲突错误是因为语言中某个地方已经存在 uint 的类型声明，并且我认为该类型是 uint_least32_t 因为:

#include <iostream>
using namespace std;

int main() {
    uint i = -1;
    cout << i << endl;
}

返回一个整数值，即 (2^32)-1。因此，是否有可能在 c++ 中将 uint 重新定义为 uint_least64_t。

Answer 1

从报错信息可以看出，在你的平台上<iostream>拉入 <string> , 拉入 <cstdlib> , 拉入 <stdlib.h> , 拉入 <sys/types.h> .

查看源代码，我们可以看到<sys/types.h>这样做:

#ifdef __USE_MISC
/* Old compatibility names for C types.  */
typedef unsigned long int ulong;
typedef unsigned short int ushort;
typedef unsigned int uint;
#endif

#ifdef __USE_MISC守卫是 feature_test_macros 的一部分系统。 __USE_MISC如果您请求 _DEFAULT_SOURCE，则在内部设置(如果您没有其他要求，也会设置)。

因此，您应该能够通过使用 -ansi 进行编译来绕过该问题。或 -std=... 之一选项(例如 -std=c++11 )。