当一个整数溢出在C前pression时会发生什么？ [英] What happens when a integer overflow occurs in a C expression?
问题描述
我有以下的C code：
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;
这是另类实现：
uint8_t firstValue = 111;
uint8_t secondValue = 145;
if (firstValue + secondValue > 0xFF) {
return true;
}
return false;
第一个例子是显而易见的， uint16_t 类型是大到足以容纳结果。
当我试图用铛在OS编译器的第二个例子/ X，它正确返回真。那里发生什么？是否有某种的临时的，更大的类型包含结果？
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?
推荐答案
+ 的操作数提升到更大的类型，我们可以去看到这个的草案C99标准部分 6.5.6 的加法运算符的它说：
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:
如果两个操作数的算术类型，通常的算术转换执行上
它们。
If both operands have arithmetic type, the usual arithmetic conversions are performed on them.
如果我们去 6.3.1.8 的常见的算术转换的它说：
另外，整数促销活动是在两个操作数执行。
Otherwise, the integer promotions are performed on both operands.
然后我们去 6.3.1.1 的布尔，字符和整数的它说（的重点煤矿的）
and then we go to 6.3.1.1 Boolean, characters, and integers which says (emphasis mine):
如果int可以重新present原始类型的所有值，该值被转换为int;
否则，将其转换为一个unsigned int。 这些被称为整数
促销活动 0.48），而所有其它类型的整数提升不会改变。
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.
所以 + 在这种情况下，两个操作数将晋升为键入的 INT 的的操作，所以没有溢出。
So both operands of + in this case will be promoted to type int for the operation, so there is no overflow.
请注意，为什么一定要短在C转换为int算术运算之前和C ++？说明理由促销活动。
Note, Why must a short be converted to an int before arithmetic operations in C and C++? explains the rationale for promotions.
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