二元操作符的返回值

今天同事遇到了个奇怪的问题，我帮忙看了一下，发现了比较神奇的事情，记录一下。

模拟一下代码基本上是这样：

int a = 10 ;
unsigned int b = 2 ;
if (b - a < 0)
{
// do something
}

然后我发现 b - a < 0这个判断返回的是false，也就是这个if语句不成立。

google了一下在stackoverflow里找到了答案，这是C的标准，当 int 和 unsigned int 操作时，int会被强制转换成unsigned int ，也就是说 b - a的结果也是一个unsigned int，它的值总是>= 0的，在XCode 5.1上这段代码直接就给出警告了（"Comparison of unsigned expression < 0 is always false"）

下面是stackoverflow的链接：

http://stackoverflow.com/questions/6770258/how-do-promotion-rules-work-when-the-signedness-on-either-side-of-a-binary-opera

引用accepted answer中的一段，讲述在C中二元操作符是如何处理两个不同类型变量的。

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

• If either operand is of type long double, the other shall be converted to long double.
• Otherwise, if either operand is double, the other shall be converted to double.
• Otherwise, if either operand is float, the other shall be converted to float.
• Otherwise, the integral promotions shall be performed on both operands.
• Then, if either operand is unsigned long the other shall be converted to unsigned long.
• Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int.
• Otherwise, if either operand is long, the other shall be converted to long.
• Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

[Note: otherwise, the only remaining case is that both operands are int]

后记：其实这些内容在深入理解计算机系统这本书里早有介绍，只能说自己的基础还是不够，惭愧了，总之还要多读书啊。