如何计算用C计算机字长？ [英] How to compute the word size of your computer in C?

问题描述

可能重复：结果
  确定我的处理器字长

这是一个今天的面试问题。但我不知道...

It is One Interview question today. But I didn't know ...

我觉得面试官meaned CPU的字长。

I think the interviewer meaned the word size of cpu.

我觉得像这样的答案：

int cpu_bits(void *dummy1, void *dummy2) 
{ 
 long offset = (long)&dummy2 - (long)&dummy1; 
 int ret = 0; 
 if (8 == offset) 
     ret = 64; 
 else if (4 == offset) 
     ret = 32; 
 else if (2 == offset) 
     ret = 16; 
 else if (1 == offset) 
     ret = 8; 
 else 
     ret = -1; 
 return ret;  
} 

int main() 
{ 
 printf("%d\n", cpu_bits(NULL, NULL)); 
 return 0; 
} 

的结果似乎是正确的，你这样认为吗？

The result seems to be right, Do you think so ?

推荐答案

简短的回答：标准没有定义是保证对应于底层架构的字大小的数据类型，以及什么叫字的大小是指现代的CPU是一个相当模糊的事情：字大小与地址大小

Short answer: the standard does not define a data type that is guaranteed to correspond to the word size of the underlying architecture, and what "word size" means on modern CPU's is quite a vague thing: Word Size versus Address Size.

对于具有兼容模式，不同大小的寄存器当前处理器，先进的解决适用于各种宽度的数据模式和指令，一般的字大小的交谈是即时通讯precise，至少可以这样说。

With current processors having compatibility modes, registers of a different size, advanced addressing modes and instructions suited for data of various widths, talking of a general "word size" is imprecise, to say the least.

我想面试官仍住在上世纪90年代，并记得是在半信半疑名为字和 DWORD 类型通过WinAPI的介绍时，大多数电脑仍然为16位。

I suppose the interviewer is still living in the 90's and remembers the dubiously called WORD and DWORD types that were introduced by WinAPI when most computers were still 16-bit.

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