RAND_MAX宏：符号或无符号？ [英] RAND_MAX macro: signed or unsigned?

问题描述

我看过了C标准（1999年），它只是说 RAND_MAX 至少应为32767，但没有谈到这个宏是否应该扩展为说签署或一个unsigned int。统一UNIX规范（链接1 ，的链接2 ）和Linux的人（的链接）不添加任何清晰。

I've looked up the C standard (from 1999) and it only says that RAND_MAX should be at least 32767 but says nothing about whether this macro should expand to a signed or an unsigned int. The Single UNIX Specification (link 1, link 2) and Linux man (link) don't add any clarity.

有人会认为 RAND_MAX 应该是符号int 因为那是兰特（） 的回报。

One would think RAND_MAX should be a signed int since that's what rand() returns.

不过，我发现，有些编译器把它定义为无符号：

However, I've found that some compilers define it as unsigned:

• 古代的Turbo C ++ 1.01：＃定义RAND_MAX 0x7FFFU


• 不那么古老的C ++ Builder的5.5：＃定义RAND_MAX 0x7FFFU


• 的还活着公开赛在Watcom C / C ++ 1.9：＃定义RAND_MAX 32767U


• DJGPP（DOS版的gcc 3.3.4）：＃定义RAND_MAX 2147483647


• MinGW的（用于Windows 4.6.2 GCC）：＃定义RAND_MAX 0x7FFF的


• MS Visual Studio 2010中（链接）：< I> RAND_MAX被定义为数值0x7FFF的的


• 微小的C编译器0.9.25：＃定义RAND_MAX 0x7FFF的


• LCC-win32的3.8：＃定义RAND_MAX 0x7FFF的


• PellesÇ6.50：＃定义RAND_MAX 0x3fffffff 或的#define RAND_MAX 0x7FFF的


• 数字火星的C / C ++ 8.52：＃定义RAND_MAX 32767

• The ancient Turbo C++ 1.01: #define RAND_MAX 0x7FFFU
• The not so ancient C++ Builder 5.5: #define RAND_MAX 0x7FFFU
• The still alive Open Watcom C/C++ 1.9: #define RAND_MAX 32767U
• DJGPP (gcc 3.3.4 for DOS): #define RAND_MAX 2147483647
• MinGW (gcc 4.6.2 for Windows): #define RAND_MAX 0x7FFF
• MS Visual Studio 2010 (link): RAND_MAX is defined as the value 0x7fff
• Tiny C Compiler 0.9.25: #define RAND_MAX 0x7FFF
• lcc-win32 3.8: #define RAND_MAX 0x7fff
• Pelles C 6.50: #define RAND_MAX 0x3fffffff OR #define RAND_MAX 0x7fff
• Digital Mars C/C++ 8.52: #define RAND_MAX 32767

这使得看似无害code像下面变成非便携和炸毁由于签署无符号的推广：

This makes seemingly innocuous code like the below become non-portable and blow up due to signed to unsigned promotion:

cos(w * t) + (rand() - RAND_MAX / 2) * 0.1 / (RAND_MAX / 2);

兰特（）返回值的范围在[0， RAND_MAX <α符号int / code>。

rand() returns a signed int in the range [0,RAND_MAX].

如果 RAND_MAX 定义为 unsigned int类型，从兰特（）的值提升为 unsigned int类型了。

If RAND_MAX is defined as an unsigned int, the value from rand() gets promoted to unsigned int too.

如果是这样的话，差异（RAND（） - RAND_MAX / 2）成为无符号整数与范围[0值的无符号的差异， RAND_MAX - RAND_MAX / 2]安培; [ UINT_MAX +1 - RAND_MAX / 2， UINT_MAX - 1]而不是被符号整数与在区间[数值签署的区别 - RAND_MAX / 2， RAND_MAX - RAND_MAX / 2]

And if that's the case, the difference (rand() - RAND_MAX / 2) becomes an unsigned difference of unsigned integers with the value in the ranges [0,RAND_MAX-RAND_MAX/2] & [UINT_MAX+1-RAND_MAX/2,UINT_MAX-1] instead of being a signed difference of signed integers with the value in the range [-RAND_MAX/2,RAND_MAX-RAND_MAX/2].

总之，它看起来像 RAND_MAX 应签署并最多（？）编译器将其定义为这样的，但有没有，说应该签署任何权威的来源？旧标准？ K＆安培; R？另一个UNIX规范？

Anyhow, it seems like RAND_MAX should be signed and most(?) compilers define it as such, but is there any authoritative source that says it should be signed? Older standard? K&R? Another UNIX spec?

推荐答案

答案是：code正在无理的假设，因此它需要修复。这是什么code应该做的是投 RAND_MAX 到（INT）使用情况。

The answer is: that code is making an unwarranted assumption, so it needs to be fixed. What that code should have done is cast RAND_MAX to (int) on usage.

虽然它将使更多的意义编译器为签订定义其 RAND_MAX 宏，标准非常谨慎，要求他们这样做。这使得它一个可移植的bug任何code盲目认为它签署。

While it would make a bit more sense for compilers to define their RAND_MAX macros as signed, the standard carefully avoids requiring them to do so. That makes it a portability bug for any code to blindly assume it to be signed.

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