许多快乐的回报 [英] Many Happy Returns

问题描述

#include< stdio.h>


int main（无效）

{

printf（" Happy）生日，Dennis Ritchie。％d today.\
"，~0& 0x1f<< 1）;

返回0;

}


-

Richard Heathfield： bi****@eton.powernet .co.uk

Usenet是一个奇怪的地方。 - Dennis M Ritchie，1999年7月29日。

C FAQ： http://www.eskimo.com/~scs/C-faq/top.html

K& R答案，C书等：< a rel =nofollowhref =http://users.powernet.co.uk/etontarget =_ blank> http://users.powernet.co.uk/eton

解决方案

Richard Heathfield< do ****** @ address.co.uk.invalid>写道：

#include< stdio.h>

int main（void）
{
printf（生日快乐） ，Dennis Ritchie。％d today.\
"，~0& 0x1f<< 1）;
返回0;
}

这个程序不会导致不确定的行为吗？


Martin

Martin Dickopp写道：

Richard Heathfield< do ****** @ address.co.uk.invalid>写道：

#include< stdio.h>

int main（void）
{
printf（生日快乐） ，Dennis Ritchie。％d today.\ n，0& 0x1f<< 1）;
返回0;
}

没有''这个程序会导致未定义的行为吗？

#include< stdio.h>


int main（void）

{

printf（生日快乐，Dennis Ritchie。％u today.\ n，〜0u& 0x1f<< 1）;

返回0;

}


-

pete

< blockquote> Richard Heathfield< do ****** @ address.co.uk.invalid>在消息新闻中写道：< bj ********** @ titan.btinternet.com> ...

#include< stdio.h>

int main（void）
{/> printf（生日快乐，Dennis Ritchie。％d today.\ n，〜0& 0x1f<< 1）;
返回0;
}

我可以看到为什么~0将为-1，因为〜反转所有位，所以你的签名

bit最终代表负数，你的值位代表1.

但是为什么0x1f等于某个-1.994942数？我在想0x1f

对于％f而言等于1.000000而对于％d仅为1，但它不会。

为什么会这样？

#include <stdio.h>

int main(void)
{
printf("Happy Birthday, Dennis Ritchie. %d today.\n", ~0&0x1f<<1);
return 0;
}

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

解决方案

Richard Heathfield <do******@address.co.uk.invalid> writes:

#include <stdio.h>

int main(void)
{
printf("Happy Birthday, Dennis Ritchie. %d today.\n", ~0&0x1f<<1);
return 0;
}

Doesn''t this program cause undefined behavior?

Martin

Martin Dickopp wrote:

Richard Heathfield <do******@address.co.uk.invalid> writes:

#include <stdio.h>

int main(void)
{
printf("Happy Birthday, Dennis Ritchie. %d today.\n", ~0&0x1f<<1);
return 0;
}

Doesn''t this program cause undefined behavior?

#include <stdio.h>

int main(void)
{
printf("Happy Birthday, Dennis Ritchie. %u today.\n", ~0u&0x1f<<1);
return 0;
}

--
pete

Richard Heathfield <do******@address.co.uk.invalid> wrote in message news:<bj**********@titan.btinternet.com>...

#include <stdio.h>

int main(void)
{
printf("Happy Birthday, Dennis Ritchie. %d today.\n", ~0&0x1f<<1);
return 0;
}

I can see why ~0 would be -1 since ~ inverts all bits, so your signed
bit ends up representing a negative and your value bits represent 1.
But why does 0x1f equal to some -1.994942 number? I was thinking 0x1f
would be equal to 1.000000 for %f and for %d just 1, but it doesn''t.
Why is this?

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