blocks|key|2879472|text|我认为你想要更接近这个效果的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2879473|remainingGuess+=+guesses;
do+{
+++System.out.println("Enter+guess+#1");
+++userGuess+=+keyboard.nextInt();
}+while+(+userGuess+!=+actualNumber+%7C%7C+remainingGuesses--+>+0+)|code-block|syntax|javascript|2879474|逐行：|2879475|remainingGuesses+=+guesses;|2879476|将guesses赋值给remainingGuesses一次，如果您在循环的每次迭代中都这样做，那么它将永远不会结束|offset|length|style|CODE|2879477|userGuess+!=+actualNumber+%7C%7C+remainingGuesses--+>+0+|2879478|当用户猜测错误或者remainingGuesses大于0时，继续迭代。|2879479|remainingGuesses--|2879480|计算为变量remainingGuesses的当前值，然后在表达式将其减去1之后。|2879481|entityMap^0|0|0|0|0|1|7|B|G|0|0|9|G|R|1|0|0|5|G|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@$M|15|N|16|O|P]|$M|17|N|18|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1A|8|@$M|1B|N|1C|O|P]|$M|1D|N|1E|O|P]]|9|@]|A|$]]|$1|U|3|V|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1G|8|@$M|1H|N|1I|O|P]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|Z|$]]

I think that you want something closer to this effect: 

<pre><code>remainingGuess = guesses;
do {
 System.out.println("Enter guess #1");
 userGuess = keyboard.nextInt();
} while ( userGuess != actualNumber || remainingGuesses-- &gt; 0 )
</code></pre>

Line by line:

<pre><code>remainingGuesses = guesses;
</code></pre>

assigns <code>guesses</code> to <code>remainingGuesses</code> once, if you were to do it every iteration of your loop, it would never end

<pre><code>userGuess != actualNumber || remainingGuesses-- &gt; 0 
</code></pre>

Keep iterating while the user has guessed incorrectly OR <code>remainingGuesses</code> is more than <code>0</code>. 

<pre><code>remainingGuesses--
</code></pre>

Evaluates to the current value of the variable <code>remainingGuesses</code>, then after the expression decrements it by 1.

blocks|key|2879483|text|在进入循环之前初始化剩余的猜测。只检查while括号"()“中的条件|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2879484|reaminingGuesses+=+guesses;
do
++++{System.out.println("Enter+guess+#1");
++++userGuess+=+keyboard.nextInt();
++++remainingGuess--;
++++}+while+((userGuess+!=+actualNumber)+%7C%7C+(remainingGuesses+>=+1));+|code-block|syntax|javascript|2879485|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|$]]

Initialize remaining guesses before entering the loop. Only check the condition in the while paranthesis "()"

<pre><code>reaminingGuesses = guesses;
do
 {System.out.println("Enter guess #1");
 userGuess = keyboard.nextInt();
 remainingGuess--;
 } while ((userGuess != actualNumber) || (remainingGuesses &gt;= 1)); 
</code></pre>

blocks|key|695998|text|如果你想这样做，我会做一些小的修改，并假设你已经为actualNumber和允许的猜测总数定义了常量：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|695999|//+break+the+calls+into+seperate+methods,+to+make+them+easier+to+read+in+the+long+term+and+to+
//+seperate+logical+concerns+into+discrete+elements.
do+{
//++get+the+user+guess
+++guess+=+getGuess();
//+if+shouldStop+returns+true+you+would+stop,+if+it+returns+false+you+should+continue,+thus+the+!..
}++while+(!shouldStop(guess))+{
+++//+decrement+the+guesses.
+++--remainingGuesses;
}

///++define+these+methods+somewhere..,
//+get+your+guess
public+int+getGuess()+{
+++System.out.println("Enter+guess+#1");
+++return+keyboard.nextInt();
}


public+boolean+shouldStop(int+guess)+{
//+condensed+if+statement+syntax+in+java,+very+handy+for+this+kind+of+thing
++++//+do+we+have+enough+remaining+guess+to+keep+going?++If+so,+compare+the+guess+to+the+actual+number
+++//+otherwise+return+true+(the+stop+signal
+++return+remainingGuesses+>+0+?+guess+==+actualNumber+:+true;
}|code-block|syntax|javascript|696000|如果你真的想要完成整个过程，你也可以将guess==actual数分解成一个方法，但在这种情况下可能不需要这样做，因为这是一个简单的相等检查。|696001|然而，methos+shouldStop可以通过多种方式定义。|offset|length|style|CODE|696002|早些时候，我认为完整地写出这些逻辑块并从中压缩是有帮助的，例如：|696003|public+boolean+shouldStop(int+guess)+{
++if(remainingGuesses+<=0)+{
+++++return+true;
++}++else+if(guess+==+actualNumber)+{
+++++return+true;
++}++else+{
+++++return+false;
}|696004|}|696005|entityMap^0|0|0|0|A|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

If you want it as a do while I would make a few minor changes, and assuming you have constants defined for things like actualNumber and the total number of allowed guesses:

<pre><code>// break the calls into seperate methods, to make them easier to read in the long term and to 
// seperate logical concerns into discrete elements.
do {
// get the user guess
 guess = getGuess();
// if shouldStop returns true you would stop, if it returns false you should continue, thus the !..
} while (!shouldStop(guess)) {
 // decrement the guesses.
 --remainingGuesses;
}

/// define these methods somewhere..,
// get your guess
public int getGuess() {
 System.out.println("Enter guess #1");
 return keyboard.nextInt();
}


public boolean shouldStop(int guess) {
// condensed if statement syntax in java, very handy for this kind of thing
 // do we have enough remaining guess to keep going? If so, compare the guess to the actual number
 // otherwise return true (the stop signal
 return remainingGuesses &gt; 0 ? guess == actualNumber : true;
}
</code></pre>

If you really wanted to follow the whole thing through you could break the guess==actual number into a method as well, but its probably not needed in this case since its a simple equality check.

The methos <code>shouldStop</code> could be defined a number of ways however...

Early on I think it's helpful to write these logical blocks out fully, and condense from there, for example:

<pre><code>public boolean shouldStop(int guess) {
 if(remainingGuesses &lt;=0) {
 return true;
 } else if(guess == actualNumber) {
 return true;
 } else {
 return false;
}
</code></pre>

}

newbie programmer here. I'm using Java to try and create a number guessing game. I want my do while loop to continue looping until the user inputs the correct number OR they run out of guesses. This is what I have and I can't figure out how to use 2 boolean controllers for the life of me.

<pre><code>do
 {System.out.println("Enter guess #1");
 userGuess = keyboard.nextInt();

 } while ((userGuess != actualNumber) || (remainingGuesses = guesses; remainingGuesses &gt;= 1; remainingGuesses--)); 
</code></pre>

Any help is appreciated!

Putting 2 Boolean controllers in a do while expression?

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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新手程序员来了。我正在使用Java来尝试创建一个猜数游戏。我希望do while循环继续循环，直到用户输入正确的数字或猜不到为止。这就是我所拥有的，我不知道如何使用2个布尔控制器。do    {System.out.println("Enter guess #1");    userGuess = keyboard.n...

问在do while表达式中放入2个布尔控制器？
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问在do while表达式中放入2个布尔控制器？EN