macos 在 64 位 CPU 上使用 C++ 编写的 Mac OS X 上，是否有 64 位类型？
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On Mac OS X in C++ on a 64-bit CPU, is there a type that is 64 bits?
提问by anon
I can't use "long long"; what should I be using?
我不能用“long long”；我应该使用什么？
回答by Jonathan Leffler
Assuming Snow Leopard (Mac OS X 10.6.2 - Intel), then 'long' is 64-bits with the default compiler.
假设 Snow Leopard（Mac OS X 10.6.2 - Intel），则“long”是 64 位，使用默认编译器。
Specify 'g++ -m64' and it will likely be 64-bits on earlier versions too.
指定 'g++ -m64' 并且在早期版本上也可能是 64 位。
1 = sizeof(char)
1 = sizeof(unsigned char)
2 = sizeof(short)
2 = sizeof(unsigned short)
4 = sizeof(int)
4 = sizeof(unsigned int)
8 = sizeof(long)
8 = sizeof(unsigned long)
4 = sizeof(float)
8 = sizeof(double)
16 = sizeof(long double)
8 = sizeof(size_t)
8 = sizeof(ptrdiff_t)
8 = sizeof(time_t)
8 = sizeof(void *)
8 = sizeof(char *)
8 = sizeof(short *)
8 = sizeof(int *)
8 = sizeof(long *)
8 = sizeof(float *)
8 = sizeof(double *)
8 = sizeof(int (*)(void))
8 = sizeof(double (*)(void))
8 = sizeof(char *(*)(void))
Tested with:
测试：
i686-apple-darwin10-g++-4.2.1 (GCC) 4.2.1 (Apple Inc. build 5646) (dot 1) Copyright (C) 2007 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
Compiling with GCC 4.7.1 on Mac OS X 10.7.5 with option -std=c99, the output from the program is more extensive. Thanks to apalopohapafor pointing out the oversight that long longetc were missing from the original.
使用选项 在 Mac OS X 10.7.5 上使用 GCC 4.7.1 进行编译，-std=c99程序的输出更广泛。感谢apalopohapa指出long long原文中缺少的疏忽。
1 = sizeof(char)
1 = sizeof(unsigned char)
2 = sizeof(short)
2 = sizeof(unsigned short)
4 = sizeof(int)
4 = sizeof(unsigned int)
8 = sizeof(long)
8 = sizeof(unsigned long)
4 = sizeof(float)
8 = sizeof(double)
16 = sizeof(long double)
8 = sizeof(size_t)
8 = sizeof(ptrdiff_t)
8 = sizeof(time_t)
8 = sizeof(long long)
8 = sizeof(unsigned long long)
8 = sizeof(uintmax_t)
1 = sizeof(int8_t)
2 = sizeof(int16_t)
4 = sizeof(int32_t)
8 = sizeof(int64_t)
1 = sizeof(int_least8_t)
2 = sizeof(int_least16_t)
4 = sizeof(int_least32_t)
8 = sizeof(int_least64_t)
1 = sizeof(int_fast8_t)
2 = sizeof(int_fast16_t)
4 = sizeof(int_fast32_t)
8 = sizeof(int_fast64_t)
8 = sizeof(uintptr_t)
8 = sizeof(void *)
8 = sizeof(char *)
8 = sizeof(short *)
8 = sizeof(int *)
8 = sizeof(long *)
8 = sizeof(float *)
8 = sizeof(double *)
8 = sizeof(int (*)(void))
8 = sizeof(double (*)(void))
8 = sizeof(char *(*)(void))
1 = sizeof(struct { char a; })
2 = sizeof(struct { short a; })
4 = sizeof(struct { int a; })
8 = sizeof(struct { long a; })
4 = sizeof(struct { float a; })
8 = sizeof(struct { double a; })
16 = sizeof(struct { char a; double b; })
16 = sizeof(struct { short a; double b; })
16 = sizeof(struct { long a; double b; })
4 = sizeof(struct { char a; char b; short c; })
16 = sizeof(struct { char a; char b; long c; })
4 = sizeof(struct { short a; short b; })
6 = sizeof(struct { char a[3]; char b[3]; })
8 = sizeof(struct { char a[3]; char b[3]; short c; })
16 = sizeof(struct { long double a; })
32 = sizeof(struct { char a; long double b; })
16 = sizeof(struct { char a; long long b; })
16 = sizeof(struct { char a; uintmax_t b; })
回答by F'x
Include <stdint.h>or <inttypes.h>(the later is found on some more compilers, but both are provided by the Apple compiler), and use uint64_tand int64_t. They are 64-bit on both 32-bit and 64-bit targets.
包括<stdint.h>or <inttypes.h>（后者在更多编译器上找到，但两者都由 Apple 编译器提供），并使用uint64_tand int64_t。它们在 32 位和 64 位目标上都是 64 位的。
