blocks|key|336667|text|我不认为这是可能的。括号里的是真或假的陈述，没有真/真或真/假这样的东西。你可以做的是一个带有break语句的do/while循环。但我不认为这是一条正确的道路。为什么要避免两个if语句？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|336668|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I don't think that's possible. What you have inside your bracket is a statement which is either true or false, there's no such thing like a true/true or true/false statement. What you could do is a do/while loop with a break statement. But I don't think that's the way to go. Why do you want to avoid two if statements?

blocks|key|4495315|text|单个"%7C“或"&”保证两侧的评估，即使结果可以由左侧运算符单独确定。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4495316|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

single "|" or "&amp;" gaurantees both side evaluation even if the result can be determined by left side operator alone.

blocks|key|1145415|text|你可以这样做，来构建“消息”：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1145416|string+msg+=+"Your+Message\n";
string+buildSt+=+x+==+y+?+m+==+n+?+msg+%2B+msg+:+msg+:+m+==+n+?+msg+:+"";+|code-block|syntax|javascript|1145417|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could do something like this, to build up the "message":

<pre><code>string msg = "Your Message\n";
string buildSt = x == y ? m == n ? msg + msg : msg : m == n ? msg : ""; 
</code></pre>

blocks|key|4977932|text|当两个条件都为真时，编译器只检查一个条件，因为你已经用OR连接了你的条件。如果ORs链中即使有一个条件为真，就不需要检查其他条件，因为结果已经为真，如果其中一个为假，则将为假。因此，如果您认为您的逻辑是正确的，那么就没有必要进行多次检查。您的代码要求您在其中一个条件为真时打印一条消息，并执行此操作。如果在两个条件都为真的情况下需要一些特殊的东西，那么可以单独添加它。简而言之，您永远不能期望编译器对由OR连接的表达式执行所有检查。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4977933|致以敬意，|4977934|戴维特|4977935|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Compiler checks only one condition when both are true because you've connected your conditions with OR.
If even one condition in ORs chain is true there is no need to check others as a result already true and will be false if one of them is false. So if you think that your logic is right then there is no need to do multiple checks. Your code is asking that you will print a message if one of the conditions is true and program doing it. If you want something special for a case when both conditions are true then add it separately. Shortly you should never expect from the compiler to do all checks in the expressions connected by OR.

Regards,

Davit

blocks|key|1145310|text|测试代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1145311|#include+<iostream>
#include+<string>
using+namespace+std;

void+main()+{
++++int+x=1;
++++int+y=1;
++++int+m=1;
++++int+n=1;
++++string+mess1="Your+message+1\n";
++++string+mess2="Your+message+2\n";
++++cout<<((x==y)?mess1:"")%2B((m==n)?mess2:"");
++++getchar();
}|code-block|syntax|javascript|1145312|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Tested code:

<pre><code>#include &lt;iostream&gt;
#include &lt;string&gt;
using namespace std;

void main() {
 int x=1;
 int y=1;
 int m=1;
 int n=1;
 string mess1="Your message 1\n";
 string mess2="Your message 2\n";
 cout&lt;&lt;((x==y)?mess1:"")+((m==n)?mess2:"");
 getchar();
}
</code></pre>

blocks|key|4495249|text|如果您正在尝试查看这两个语句是否都正确，则需要使用&&。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4495250|在比较布尔值(true/false)时，请查看Boolean+Operators以了解所有可能的选项。|4495251|回答你的问题：|4495252|++++if+((x==y)+&&+(m==n))
++++{
++++++++cout<<"Your+Message"<<endl<<"Your+Message"<<endl;
++++}
++++else+if((x==y)+%7C%7C+(m==n))++
++++{
++++++++cout<<"Your+Message"<<endl;
++++}|code-block|syntax|javascript|4495253|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B^0|P|2|0|N|H|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@$9|10|A|11|1|12]]|E|$]]|$1|H|3|I|5|6|7|13|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|14|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

If you are trying to see if both statements are true an <code>&amp;&amp;</code> is what you will want to use.

Take a look at <a href="http://en.wikipedia.org/wiki/Operators_in_C_and_C++" rel="nofollow">Boolean Operators</a> to see all of the possible options when comparing boolean (true/false) values.

To answer your question:

<pre><code> if ((x==y) &amp;&amp; (m==n))
 {
 cout&lt;&lt;"Your Message"&lt;&lt;endl&lt;&lt;"Your Message"&lt;&lt;endl;
 }
 else if((x==y) || (m==n)) 
 {
 cout&lt;&lt;"Your Message"&lt;&lt;endl;
 }
</code></pre>

I have the following C++ code:

<pre><code>if(x==y||m==n){
cout&lt;&lt;"Your message"&lt;&lt;endl;
}
</code></pre>

If x is equal to y or m is equal to n, the program prints "Your message". But if both conditions are true,the program tests only one of them and eventually prints one "Your Message". 

Is there a way to print each <code>"Your message"</code> independently based on each condition using a single if statement?

The output would be identical to the below using multiple if statements.

<pre><code>if(x==y){
cout&lt;&lt;"Your message"&lt;&lt;endl;
}

if (m==n){
cout&lt;&lt;"Your message"&lt;&lt;endl;
}
</code></pre>

Check multiple OR operators in IF statement

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解答不清晰，无法理解或解决问题。

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我有以下C++代码：if(x==y||m==n){cout<<"Your message"<<endl;}如果x等于y或m等于n，程序会打印"Your message“。但如果这两个条件都为真，程序将只测试其中一个条件，并最终打印一条“您的消息”。有没有一种方法可以使用一条if语句根据每个条件独立地打印每个"Your message"？如果使用多个if语句，输出将与以下内容相同。if(x==y){

问检查IF语句中的多个OR运算符
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