（y> = 10）|| （x ++> 10）为什么x = 10？ [英] (y >= 10) || (x++ > 10) why x=10 here ?

问题描述

假设x = 10且y = 10，在评估表达式（y> = 10）||后，x是什么（x ++> 10）。



我在c ++编译器中试过它，答案是= 10;但是x的期望值是11。

你能给我10个简单跟踪的原因

解决方案

有一个功能称为短路评估 [ ^ ]在某些语言中。因此，如果您使用副作用 [ ^ ]在布尔条件下，您可以获得一些意想不到的乐趣。



在你的情况下会发生这种情况：



OR表达式的左项：（y> = 10）

OR的右项-expression：（x ++> 10）



如果左项已经为真，则甚至不必评估OR表达式的右手项真的是|| true与true ||相同假。所以这没关系，这将是一个多余的操作。因此，在这种情况下永远不会评估 x ++ ，因此x仍然是10。 ：D

如果和表达式的左侧是假的，那么对于AND表达式也是如此，因为false&& true和false一样假，&&假。



当人们知道短路评估时，很容易回答。人们应该意识到，将副作用与布尔表达式混合可能会让您稍后进行一些讨论。 ; P







问候，

- Manfred

这是一个有趣的...





在除下文外，主要原因是短路评估。请参阅解决方案2.





有什么区别：

 x ++ 

和



< pre lang =C ++> ++ x

？



第一个返回值FIRST，然后递增x。所以如果我写的：

  int  x =  10 ; 
  int  y = x ++; 

和看完之后评价，x为11，y为10.



或者，如果我写的：

  int  x =  10 ; 
  int  y = ++ x; 

评估后，x和y都是11.



所以在你的情况下：

  if （（y> =  10 ）||（x ++>  10 ）） //  评估期间x为10，之后为11。  
 {
 
} 

因为你没有全线，我只能假设这是if语句。



如果你写了类似的东西：

  int  y =  9 ; 
  int  x =  10 ; 
  if （（y> =  10 ）||（x ++>  10 ）||（x>  10 ））
 {
 
} 

第一个（x ++> 10）将评估为false，但（x> 10）将评估为true。

在你的情况下，第一个条件本身是真的，所以||操作员右侧表达式不会被执行。

Suppose x=10 and y=10 what is x after evaluating the expression (y >= 10) || (x++ > 10).

i tried it in c++ compiler and the answer is = 10; but the expected value of x to be 11 .
could you give me the reasons why 10 with simple trace

解决方案

There is a feature called short circuit evaluation[^] in some languages. So if you are using side effects[^] in boolean conditions you are in for some unexpected fun.

In your case this happens:

Left term of OR-expression: (y >= 10)
Right term of OR-expression: (x++ > 10)

If the left term is already true, one doesn't even have to evaluate the right hand term of the OR-expression since true || true is true the same as true || false. So it doesn't matter and it would be a superfluous operation. Thus x++ is never evaluated in this scenario and x will thus still be ten. :D
The same holds true for AND expressions if the left-hand side of the and expression is false there's no need to evaluate the right-hand side since false && true is false the same as false && false.

So it's quite easy to answer when one knows about short circuit evaluation. One should be aware that mixing side effects with boolean expressions may introduce you to some headscratching later on. ;P



Regards,

— Manfred

This is a fun one...

[Edit]
In addition to the below, the main cause is short circuit evaluation. See Solution 2.


What is the difference between:

x++

and

++x

?

The first returns the value FIRST, then increments x. So if I wrote:

int x = 10;
int y = x++;

and after looking at the evaluation, x would be 11, and y would be 10.

Alternatively, if I wrote:

int x = 10;
int y = ++x;

after evaluation, x and y would both be 11.

So in your case:

if ((y >= 10) || (x++ > 10)) //x is 10 during the evaluation and 11 after.
{

}

Since you don't give the whole line, I can only assume that this is in an if statement.

If you wrote something like:

int y = 9;
int x = 10;
if ((y >= 10) || (x++ > 10) || (x > 10))
{

}

The first (x++ > 10) would evaluate to false, but (x > 10) would evaluate to true.

In your case first condition itself is true so || operator right side expression will not be executed.

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