C++ unsigned long long 与 uint64_t 冲突？
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/32198368/
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
unsigned long long conflict with uint64_t?
提问by thundium
We use template specialization for some type parameter like
我们对某些类型参数使用模板特化，例如
class my_template_class<uint64_t M>: public my_template_class_base<uint64_t> {
....
}
class my_template_class<unsigned long long,M>: public my_template_class_base<unsigned long long> {
....
}
This is working perfectly with 64-bit compilation with gcc. While when we try the 32 bit mode, it reports "previous definition" for above two classes.
这与 gcc 的 64 位编译完美配合。而当我们尝试 32 位模式时，它会报告上述两个类的“先前定义”。
So unsigned long longis the same as uint64_tin the 32-bit compilation but not in 64-bit compliation?
那么在32位编译中unsigned long long是否相同uint64_t但在64位编译中不同？
The compilation difference is the CXXflag -m32and -m64
编译区别在于CXX标志-m32和-m64
回答by Keith Thompson
So
unsigned long longis the same asuint64_tin the 32-bit compilation but not in 64-bit compilation?
那么
unsigned long long是否与uint64_t32 位编译中相同但在 64 位编译中不同？
Yes.
是的。
In 32-bit mode, most likely longis 32 bits and long longis 64 bits. In 64-bit mode, both are probably 64 bits.
在 32 位模式下，最有可能long是 32 位和long long64 位。在 64 位模式下，两者都可能是 64 位。
In 32-bit mode, the compiler (more precisely the <stdint.h>header) defines uint64_tas unsigned long long, because unsigned longisn't wide enough.
在 32 位模式下，编译器（更准确地说是<stdint.h>头文件）定义uint64_t为unsigned long long，因为unsigned long不够宽。
In 64-bit mode, it defines uint64_tas unsigned long.
在 64 位模式下，它定义uint64_t为unsigned long.
It couldhave defined it as unsigned long longin both modes. The choice is arbitrary; all that's required is that it has to be a 64-bit type.
它可以将其定义为unsigned long long两种模式。选择是任意的；所需要的只是它必须是 64 位类型。
In general, each of the integer types defined in <stdint.h>is a typedef for some predefined type with the appropriate characteristics. You can't assume that any of them are distinct from the predefined types.
通常，定义的每个整数类型<stdint.h>都是具有适当特征的某些预定义类型的 typedef。您不能假设它们中的任何一个都不同于预定义的类型。
回答by Paolo M
This is from stdint.hfor GCC 4.8:
这是来自stdint.hGCC 4.8：
#if __WORDSIZE == 64
typedef unsigned long int uint64_t;
#else
__extension__
typedef unsigned long long int uint64_t;
#endif
So:
所以：
So unsigned long long is the same as uint64_t in the 32bit compliation but not in 64 bit compliation?
所以 unsigned long long 和 uint64_t 在 32 位编译中是一样的，但在 64 位编译中不一样？
Yes.
是的。
