C++ 为什么数字限制对 uint8_t 和 int8_t 不起作用?

Par*_*dox 2 c++ numeric-limits

我最近注意到 numeric_limits::max() 和 numeric_limits::min() 似乎不适用于 uint8_t 和 int8_t。这是有原因的还是可能是错误?我在自己的电脑上使用 gcc 编译器尝试过:

#include <iostream>
#include <limits>

using namespace std;

int main()
{

    std::cout << "numeric_limits<uint8_t>::max() = " << numeric_limits<uint8_t>::max() << std::endl; 
    std::cout << "numeric_limits<int8_t>::max() = " << numeric_limits<int8_t>::max() << std::endl; 
    std::cout << "numeric_limits<int8_t>::min() = " << numeric_limits<int8_t>::min() << std::endl;

    std::cout << "numeric_limits<uint16_t>::max() = " << numeric_limits<uint16_t>::max() << std::endl; 
    std::cout << "numeric_limits<int16_t>::max() = " << numeric_limits<int16_t>::max() << std::endl; 
    std::cout << "numeric_limits<int16_t>::min() = " << numeric_limits<int16_t>::min() << std::endl; 

    std::cout << "numeric_limits<uint32_t>::max() = " << numeric_limits<uint32_t>::max() << std::endl; 
    std::cout << "numeric_limits<int32_t>::max() = " << numeric_limits<int32_t>::max() << std::endl; 
    std::cout << "numeric_limits<int32_t>::min() = " << numeric_limits<int32_t>::min() << std::endl; 

    std::cout << "numeric_limits<uint64_t>::max() = " << numeric_limits<uint64_t>::max() << std::endl; 
    std::cout << "numeric_limits<int64_t>::max() = " << numeric_limits<int64_t>::max() << std::endl; 
    std::cout << "numeric_limits<int64_t>::min() = " << numeric_limits<int64_t>::min() << std::endl; 

    return 0;
}
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给出输出:

numeric_limits<uint8_t>::max() = ?
numeric_limits<int8_t>::max() = 
numeric_limits<int8_t>::min() = ?
numeric_limits<uint16_t>::max() = 65535
numeric_limits<int16_t>::max() = 32767
numeric_limits<int16_t>::min() = -32768
numeric_limits<uint32_t>::max() = 4294967295
numeric_limits<int32_t>::max() = 2147483647
numeric_limits<int32_t>::min() = -2147483648
numeric_limits<uint64_t>::max() = 18446744073709551615
numeric_limits<int64_t>::max() = 9223372036854775807
numeric_limits<int64_t>::min() = -9223372036854775808
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Mik*_*l H 6

它确实有效。但是,输出被解释为 ASCII 字符。如果在打印之前转换为 int,您将看到正确的值:

std::cout << "numeric_limits<uint8_t>::max() = " << static_cast<int>(numeric_limits<uint8_t>::max()) << std::endl; 
std::cout << "numeric_limits<int8_t>::max() = " << static_cast<int>(numeric_limits<int8_t>::max()) << std::endl; 
std::cout << "numeric_limits<int8_t>::min() = " << static_cast<int>(numeric_limits<int8_t>::min()) << std::endl;
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