blocks|key|1080214|text|问题是您有一个后跟else+if的if，但是如果这两个检查都失败了，那么默认情况下将不会返回任何内容。您的代码应该防止这种情况发生，但是编译器不够聪明，无法100%25确定您永远不会在两个if检查中都失败。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1080215|try{
++++cout+<<+"Please+enter+a+integer+between+1+and+10:+";
++++cin+>>+Num;
++++if+((Num+>+10)+%7C%7C+(Num+<+0)){
++++++++throw+77;
++++}
++++else+if+((Num+<=+10)+%7C%7C+(Num+>=+0)){
++++++++cout+<<+Num+<<+"+is+in+range\n";
++++++++return+0;
++++++++system("pause");
++++}
++++//+<-+if+you+get+here,+nothing+is+going+to+be+returned
+}|code-block|syntax|javascript|1080216|另一种方法可能是将else+if更改为只使用else|1080217|entityMap^0|9|7|H|2|2K|2|0|0|9|7|M|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

The problem is that you have an <code>if</code> followed by an <code>else if</code>, but if both of these checks fail, there's nothing to return by default. Your code should prevent that from ever happening, but the compiler isn't smart enough to know with 100% certainty that you will never fail both <code>if</code> checks.

<pre><code>try{
 cout &lt;&lt; "Please enter a integer between 1 and 10: ";
 cin &gt;&gt; Num;
 if ((Num &gt; 10) || (Num &lt; 0)){
 throw 77;
 }
 else if ((Num &lt;= 10) || (Num &gt;= 0)){
 cout &lt;&lt; Num &lt;&lt; " is in range\n";
 return 0;
 system("pause");
 }
 // &lt;- if you get here, nothing is going to be returned
 }
</code></pre>

An alternative approach might be to change your <code>else if</code> to just <code>else</code>

blocks|key|3036717|text|所以，简单解释一下:当你有一个需要返回值的函数(在你的例子中，是一个int)，当你的程序中的所有路径都这样做时，编译器会很高兴。在这段代码中，可能会出现这样一种情况，即它可能会越过您的if和else+if，在这种情况下，它不会做任何事情。如果你愿意，你可以在这里抛出一个错误或者别的什么。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3036718|但我还想添加一件额外的事情，那就是在接收用户输入时添加一些保护。也就是说，确保用户输入了一个数字！|3036719|例如，目前，如果用户输入b，则输出将是0+is+in+range！|3036720|因此，添加另一个检查以确保：|3036721|int+using_range()
{
++++int+Num;

++++try{
++++++++cout+<<+"Please+enter+a+integer+between+1+and+10:+";
++++++++cin+>>+Num;

++++++++//+this+checks+if+whatever+the+user+input+can+be+parsed+as+an+int.
++++++++if+(!cin){
+++++++++++cout+<<+"please+enter+a+number."+<<+endl;
+++++++++++return+-1;
++++++++}
++++++++else+if+((Num+>+10)+%7C%7C+(Num+<+0)){
++++++++++++throw+77;
++++++++}
++++++++else+if+((Num+<=+10)+%7C%7C+(Num+>=+0)){
++++++++++++cout+<<+Num+<<+"+is+in+range\n";
++++++++++++return+0;
++++++++++++system("pause");+//+also,+this+will+never+happen+:)
++++++++}
++++++++else{
++++++++++++//+not+sure+how+it+could+get+here,+but+maybe+in+weird+corner+cases?
++++++++++++return+-1;
++++++++}
++++}
++++catch+(int+x){
++++++++cout+<<+"The+number+cannot+be+greater+than+10,+or+less+than+zero.+Error+"+<<+x+<<+endl;
++++++++system("pause");
++++++++return+0;
++++}
}|code-block|syntax|javascript|3036722|entityMap^0|Y|3|2K|2|2N|7|0|0|C|1|J|D|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|16|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

So, just a quick explanation: When you have a function that expects a returned value (in your case, an <code>int</code>), the compiler is happy when all paths in your program actually do so. In this code, there may be a situation where it could land past your <code>if</code> and <code>else if</code>, in which case, it doesn't do anything. You could throw an error here or something if you wanted.

But one extra thing I wanted to add is to add some protection when receiving user input. That is, make sure the user entered a number!

Currently, if the user entered <code>b</code> for instance, the output would be <code>0 is in range</code>!

So, add another check to be sure:

<pre><code>int using_range()
{
 int Num;

 try{
 cout &lt;&lt; "Please enter a integer between 1 and 10: ";
 cin &gt;&gt; Num;

 // this checks if whatever the user input can be parsed as an int.
 if (!cin){
 cout &lt;&lt; "please enter a number." &lt;&lt; endl;
 return -1;
 }
 else if ((Num &gt; 10) || (Num &lt; 0)){
 throw 77;
 }
 else if ((Num &lt;= 10) || (Num &gt;= 0)){
 cout &lt;&lt; Num &lt;&lt; " is in range\n";
 return 0;
 system("pause"); // also, this will never happen :)
 }
 else{
 // not sure how it could get here, but maybe in weird corner cases?
 return -1;
 }
 }
 catch (int x){
 cout &lt;&lt; "The number cannot be greater than 10, or less than zero. Error " &lt;&lt; x &lt;&lt; endl;
 system("pause");
 return 0;
 }
}
</code></pre>

Its giving me an error saying not all controll paths return a value: please help. 

<pre><code>int using_range()
{
 int Num;

 try{
 cout &lt;&lt; "Please enter a integer between 1 and 10: ";
 cin &gt;&gt; Num;
 if ((Num &gt; 10) || (Num &lt; 0)){
 throw 77;
 }
 else if ((Num &lt;= 10) || (Num &gt;= 0)){
 cout &lt;&lt; Num &lt;&lt; " is in range\n";
 return 0;
 system("pause");
 }
 }
 catch (int x){
 cout &lt;&lt; "The number cannot be greater than 10, or less than zero. Error " &lt;&lt; x &lt;&lt; endl;
 system("pause");
 return 0;
 }
}
</code></pre>

Im not sure what to do

control paths not returning a value 3

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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它给了我一个错误，说并不是所有的控制路径都返回值:请帮助。int using_range(){ int Num; try{ cout << "Please enter a integer between 1 and 10: "; cin >> Num; if ((N...

问控制路径不返回值3
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问控制路径不返回值3EN