Question

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？我在自己的计算机上使用gcc编译器进行了尝试：

#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

Answer 1

它确实起作用。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，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;

Answer 2

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

在将它们插入到cout中之前，尝试将它们转换为字符串。

Answer 3

int8类型可能定义为char，所以不要将值打印为char而是int：

int main() {

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

C ++为什么uint8_t和int8_t的数字限制不起作用？

3 个答案: