c++ - C++为什么数字限制对uint8_t和int8_t不起作用？

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cout not printing unsigned char

(5个答案)


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我最近注意到，numeric_limits::max()和numeric_limits::min()似乎不适用于uint8_t和int8_t。是否有这个原因，或者可能是错误？我使用gcc编译器在自己的计算机上尝试过:

#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;
}

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

最佳答案

确实有效。输出虽然被解释为ASCII字符。如果在打印之前将其强制转换为int，则将看到正确的值:

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;

关于c++ - C++为什么数字限制对uint8_t和int8_t不起作用？ ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/62405311/

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Answer 1

确实有效。输出虽然被解释为ASCII字符。如果在打印之前将其强制转换为int，则将看到正确的值:

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;