对照 [英] comparison
问题描述
我想写一些像
status_bit =（unsigned）（i - j）> 31
作为
if（i< j）{
{
status_bit = 1;
}否则{
status_bit = 0;
}
这实际上适用于大多数情况，但两个值都失败
i = 0x80000000
j = 0x7fffffff
我想要一个上面的线性代码，即没有跳转，所以
1.什么可能是问题？
2.可能是什么解决方案？
I want to write something like
status_bit = (unsigned)(i - j) >31
as an equivalent code for
if(i < j){
{
status_bit = 1;
}else{
status_bit = 0;
}
This actually works for most of the cases but fails for two values
i = 0x80000000
j = 0x7fffffff
I want a linear code for the above i.e. without jumps, so
1. what could be the issue?
2. what could be the solution?
推荐答案
10月21日，3：10 pm，ndu ... @ gmail.com写道：
On Oct 21, 3:10 pm, ndu...@gmail.com wrote:
我想写类似
status_bit =（unsigned）（i - j）> 31
作为
的等价代码
I want to write something like
status_bit = (unsigned)(i - j) >31
as an equivalent code for
签署int i;
signed int j;
signed int i;
signed int j;
if（i< j）{
{
status_bit = 1;
}否则{
status_bit = 0;
}
这实际上适用于大多数但两个值都失败
i = 0x80000000
j = 0x7fffffff
我想要一个上面的线性代码，即没有跳转，所以
1.可能是什么问题？
2.可能是什么解决方案？
if(i < j){
{
status_bit = 1;
}else{
status_bit = 0;
}
This actually works for most of the cases but fails for two values
i = 0x80000000
j = 0x7fffffff
I want a linear code for the above i.e. without jumps, so
1. what could be the issue?
2. what could be the solution?
nd****@gmail.com 写道：
我想写类似
status_bit =（unsigned）（i - j）> 31 <
作为
的等价代码if（i< j）{
{
status_bit = 1;
}否则{
status_bit = 0;
}
I want to write something like
status_bit = (unsigned)(i - j) >31
as an equivalent code for
if(i < j){
{
status_bit = 1;
}else{
status_bit = 0;
}
尝试简单的
status_bit =（i< j）;
Try the simple
status_bit = (i < j);
这实际上适用于大多数情况但是两个值都失败了
i = 0x80000000
j = 0x7fffffff
我想要一个上面的线性代码，即没有跳转，所以
1.可能是什么问题？
2.可能是什么解决方案？
This actually works for most of the cases but fails for two values
i = 0x80000000
j = 0x7fffffff
I want a linear code for the above i.e. without jumps, so
1. what could be the issue?
2. what could be the solution?
是否有理由避免明显的
status_bit =（i< j）;
？
Is there a reason for avoiding the obvious
status_bit = (i < j);
?
" Martin Ambuhl" < ma ***** @earthlink.netaécritdansle message de news：
5o ************ @ mid.individual.net ...
"Martin Ambuhl" <ma*****@earthlink.neta écrit dans le message de news:
5o************@mid.individual.net...
nd **** @ gmail.com 写道：
>我想写类似
的内容status_bit =（unsigned）（i - j）> 31
作为
的等效代码if（i< j）{
{
status_bit = 1 ;
}否则{
status_bit = 0;
}
>I want to write something like
status_bit = (unsigned)(i - j) >31
as an equivalent code for
if(i < j){
{
status_bit = 1;
}else{
status_bit = 0;
}
尝试简单
status_bit = （i< j）;
Try the simple
status_bit = (i < j);
>这实际上适用于大多数情况但是两个值都失败
i = 0x80000000
j = 0x7fffffff
我想要一个上面的线性代码，即没有跳跃，所以
1。可能是什么问题？
2。可能是什么解决方案？
>This actually works for most of the cases but fails for two values
i = 0x80000000
j = 0x7fffffff
I want a linear code for the above i.e. without jumps, so
1. what could be the issue?
2. what could be the solution?
是否有理由避免明显的
status_bit =（i< j）;
Is there a reason for avoiding the obvious
status_bit = (i < j);
换句话说：如果有办法在没有测试和跳转的情况下计算这个，那么编译器应该知道
。
-
Chqrlie。
In other words: if there is a way to compute this without tests and jumps,
the compiler should know.
--
Chqrlie.
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