预期溢出但添加两个字节不会

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谁能解释一下将 uint8_t 提升为 int 的原因？这种行为的用例是什么？我可以阻止推广吗？

我有这些非常简单的代码片段，它们将两个整数类型相加并打印结果。我编译了那些 没有任何优化 和 g++ --std=c++11 main.cpp -o test

int main()
{
    uint8_t a = 200, b = 100;
    uint16_t result = (a + b);

    printf("result is %d\n", result);
}

输出：

result is 300

我原以为 (a + b) 的计算结果为 44，因为它们都是 8 位类型，应该会溢出。但是我得到了 300.

的意外结果

如果我用 uint32_t 和 uint64_t 重新运行相同的测试，它会按预期溢出：

int main()
{
    uint32_t a = UINT_MAX, b = UINT_MAX;
    uint64_t result = (a + b);

    printf("result is %ld\n", result);
}

输出：

result is 4294967294

现在我不知道为什么 uint8_t 与 uint32_t 受到不同对待。

Answer 1

小于 int 的整数类型在对其执行操作时被提升为 int

Answer 2

C++ 标准： 5.10

— Otherwise, the integral promotions (4.5) shall be performed on both operands. Then the following rules shall be applied to the promoted operands: — If both operands have the same type, no further conversion is needed.

4.5.1 积分促销

A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (4.13) is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int.

最有可能的原因是：

3.9.1 基本类型

Plain ints have the natural size suggested by the architecture of the execution environment; the other signed integer types are provided to meet special needs.

预期溢出但添加两个字节不会

Overflow expected but adding two bytes does not

c++

integer-overflow

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