How to compute the word size of your computer in C? [duplicate]
Possible Duplicate:
Determine word size of my processor
It is One Interview question today. But I didn't know ...
I think the interviewer meaned the word size of cpu.
I find an answer 开发者_开发百科like this:
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;
}
The result seems to be right, Do you think so ?
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.
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.
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.
I think they were expecting something like this:
printf("%d\n", (int)sizeof(int) * CHAR_BIT);
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