C ++中的无符号整数 [英] Unsigned integer in C++

问题描述

我编写以下代码:

 #include <iostream>
using namespace std;
int main() {

   unsigned int i=1;
   i=i-3;
   cout<<i;
   return 0;
}

输出是一个垃圾值，这是可以理解的.
现在，我编写以下代码:

The output is a garbage value, which is understandable.
Now I write the following code:

    #include <iostream>
    using namespace std;
    int main() {

    unsigned int i=1;
    i=i-3;
    i=i+5;
    cout<<i;
    return 0;
}

现在输出为3.这是怎么回事?这里的垃圾值是如何加5的?

Now the output is 3. What's happening here? How is the garbage value being added by 5 here?

推荐答案

在大钟面上绘制 unsigned int 的值的想法，最大可能值(UINT_MAX)接近零.

Think of the values of unsigned int being drawn on a large clock face with the largest possible value (UINT_MAX) being next to zero.

从1减去3将使您向后移动3个位置(这给您提供UINT_MAX-1)，向此位置加5将使您向前移动5个位置.

Subtracting 3 from 1 moves you 3 places back on the clock (which gives you UINT_MAX - 1), and adding 5 to this moves you 5 places forward.

净效果是将2加1，但重要的是要知道 intermediate 值已由C ++标准完美定义.它不是不是垃圾，而是与您平台上的 UINT_MAX 的值有关.

The net effect is to add 2 to 1, but it's important to know that the intermediate value is perfectly well defined by the C++ standard. It is not garbage, but related to the value of UINT_MAX on your platform.

请注意，此溢出的明确定义的性质对于 signed 类型而言不是正确的.在C ++中，溢出 signed 类型的行为是 undefined .

Note that the well-defined nature of this overflow is not true for signed types. The behaviour on overflowing a signed type is undefined in C++.

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