C++ 为什么数字限制不适用于uint8_t和int8_t？

C++ Why numeric limits doesn't work for uint8_t and int8_t?

本文关键字：int8 uint8 适用于 数字 为什么 不适用 C++      更新时间：2023-10-16

我最近注意到numeric_limits：：max((和numeric_limits：：min((似乎不适用于uint8_t和int8_t。这有什么原因还是可能是错误？我在自己的计算机上尝试使用gcc编译器：

#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

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;

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;

尝试将它们转换为字符串，然后再将它们插入 COUT。

int8类型可能定义为chars，因此不要将值打印为char而是打印为ints：

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;
}
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