blocks|key|3454442|text|在大多数平台上，无符号整数和有符号整数之间的区别是符号位...除此之外，实际的二进制值是相同的，它只是根据二进制值所代表的类型的符号，以两种不同的方式解释。因此，您可以明确地将负值输入“表示”为无符号值...它不会被重新解释为该值，但它肯定可以作为该值输入，而不会在流上设置错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3454443|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|$]]

The difference between unsigned and signed integers on most platforms is a sign-bit ... other than that, the actual binary value is the same, it's just interpreted two different ways depending on the sign-ness of the type that the binary value represents. So you can definitely "represent" a negative value input as a unsigned value ... it won't be re-interpreted as that value, but it can definitely be input as that value without an error being set on the stream.

blocks|key|5093931|text|不是c%2B%2B专家，但是，您是否尝试过使用cin.fail()而不是!cin，并使用cin.clear()+deeper+explaination清理缓冲区|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5093932|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/forum/beginner/2957/^0|J|A|14|B|1G|J|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Not a c++ guru or anything but, have you tried using <code>cin.fail()</code> instead of !cin and clear out your buffer with <code>cin.clear()</code>
<a href="http://www.cplusplus.com/forum/beginner/2957/" rel="nofollow">deeper explaination</a>

blocks|key|5184299|text|C和C%2B%2B允许您将负值赋给无符号类型的对象。结果是原始值减去模2%5En，其中n是无符号类型的大小。例如，unsigned+i+=+-1;将i初始化为UINT_MAX。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5184300|entityMap^0|V|3|11|1|1F|G|1W|1|21|8|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

C and C++ allow you to assign a negative value to an object of an unsigned type. The result is the original value reduced modulo <code>2^n</code>, where <code>n</code> is the size of the unsigned type. So, for example, <code>unsigned i = -1;</code> initializes <code>i</code> to <code>UINT_MAX</code>.

blocks|key|2971130|text|检查if(!cin)将仅指示未读取任何内容，例如文件的末尾。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2971131|unsigned整数并不强制输入为正；它只是意味着该值将始终被解释为正(请参阅sign+bit)，如果该数字实际上为负，则会产生显著的效果。|2971132|最好的办法可能是输入一个带符号的整数，然后在代码中测试它是否为正。|2971133|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Sign_bit^0|2|8|0|0|8|13|8|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|H|3|I|5|6|7|10|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

The check <code>if(!cin)</code> will only indicate that nothing was read, like the end of a file.

An <code>unsigned</code> integer doesn't force the input to be positive; it just means that the value will always be interpreted as positive (see <a href="http://en.wikipedia.org/wiki/Sign_bit" rel="nofollow">sign bit</a>), which can have dramatic effects if the number is in fact negative.

Your best bet is probably to input a signed integer and then test in your code whether it is positive.

blocks|key|5094038|text|或者你可以使用一个更大的有符号整型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5094039|long+long+n+=+-1;
cout+<<+"Enter+a+number:+";
cin+>>+n;

if(+!cin.good()+)
++++cout+<<+"Not+a+valid+number."+<<+endl;
else+if(+n+>+UINT_MAX+)
++++cout+<<+"Overflow,+value+is+more+than+UINT_MAX."+<<+endl;
else+if(+n+<+0+)
++++cout+<<+"Negative,+value+is+less+than+0."+<<+endl;
else+{
++++unsigned+int+m+=+(unsigned+int)n;
++++cout+<<+"Valid+unsigned+int+was+input:+"+<<+m+<<+"."+<<+endl;
}|code-block|syntax|javascript|5094040|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|$]]

Or you could just use a larger signed integral type. 

<pre><code>long long n = -1;
cout &lt;&lt; "Enter a number: ";
cin &gt;&gt; n;

if( !cin.good() )
 cout &lt;&lt; "Not a valid number." &lt;&lt; endl;
else if( n &gt; UINT_MAX )
 cout &lt;&lt; "Overflow, value is more than UINT_MAX." &lt;&lt; endl;
else if( n &lt; 0 )
 cout &lt;&lt; "Negative, value is less than 0." &lt;&lt; endl;
else {
 unsigned int m = (unsigned int)n;
 cout &lt;&lt; "Valid unsigned int was input: " &lt;&lt; m &lt;&lt; "." &lt;&lt; endl;
}
</code></pre>

I have the following program written in C++:

The problem with this program is that if the user inputs a negative number, it is NOT caught by the line <code>if(!cin)</code>. I thought that unsigned integers can NOT accept negative numbers. Then why if I enter a negative number to size, it is NOT caught by <code>if(!cin)</code> and the program continues execution with no error messages?

I cannot make use of <code>if(size &lt; 0)</code>. I want to show how unsigned integers can solve the problem of negative input.

C++ - Unsigned integers

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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我有以下用C++编写的程序：这个程序的问题是，如果用户输入一个负数，它不会被if(!cin)行捕获。我以为无符号整数是不能接受负数的。那么，为什么如果我输入一个负数来调整大小，它不会被if(!cin)捕获，程序会继续执行而不会出现错误消息？我不能使用if(size < 0)。我想展示无符号整数是如何解决负输入问题的。

问C++ -无符号整数
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回答 5

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