在C表达式中发生整数溢出时会发生什么？(What happens when a integer overflow occurs in a C expression?)

在C表达式中发生整数溢出时会发生什么？(What happens when a integer overflow occurs in a C expression?)

我有以下C代码：

uint8_t firstValue = 111; uint8_t secondValue = 145; uint16_t temp = firstValue + secondValue; if (temp > 0xFF) { return true; } return false;

这是替代实现：

uint8_t firstValue = 111; uint8_t secondValue = 145; if (firstValue + secondValue > 0xFF) { return true; } return false;

第一个例子很明显， uint16_t类型足够大以包含结果。 当我在OS / X上使用clang编译器尝试第二个示例时，它正确地返回true。 那里会发生什么？ 是否有某种临时的 ，更大的类型来包含结果？

I have the following C code:

uint8_t firstValue = 111; uint8_t secondValue = 145; uint16_t temp = firstValue + secondValue; if (temp > 0xFF) { return true; } return false;

This is the alternative implementation:

uint8_t firstValue = 111; uint8_t secondValue = 145; if (firstValue + secondValue > 0xFF) { return true; } return false;

The first example is obvious, the uint16_t type is big enough to contain the result. When I tried the second example with the clang compiler on OS/X, it correctly returned true. What happens there? Is there some sort of temporary, bigger type to contain the result?

最满意答案

+的操作数被提升为更大的类型，我们可以通过起草C99标准部分6.5.6来看到这一点。 添加操作符说：

如果两个操作数都具有算术类型，则通常会对它们执行算术转换。

如果我们转到6.3.1.8 通常的算术转换，它会说：

否则，整数升级在两个操作数上执行。

然后我们转到6.3.1.1 布尔，字符和整数 ，它们表示（ 强调我的 ）：

如果int可以表示原始类型的所有值，则该值将转换为int; 否则，它被转换为一个unsigned int。 这些被称为整数促销 .48）所有其他类型均不受整数促销的影响。

因此，在这种情况下+两个操作数将被提升为操作的int类型，所以没有溢出。

注意， 在C和C ++的算术运算之前 ， 为什么必须将一个短整数转换为整数？ 解释促销的理由。

The operands of + are promoted to larger types, we can see this by going to draft C99 standard section 6.5.6 Additive operators which says:

If both operands have arithmetic type, the usual arithmetic conversions are performed on them.

and if we go to 6.3.1.8 Usual arithmetic conversions it says:

Otherwise, the integer promotions are performed on both operands.

and then we go to 6.3.1.1 Boolean, characters, and integers which says (emphasis mine):

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.48) All other types are unchanged by the integer promotions.

So both operands of + in this case will be promoted to type int for the operation, so there is no overflow.

Note, Why must a short be converted to an int before arithmetic operations in C and C++? explains the rationale for promotions.