为什么在uint8_t
的情况下following code失败?
#include <iostream>
#include <cstdint>
#include <stack>
template <typename TT>
void PrintNumberScientificNotation (TT number) {
constexpr TT kBase{10}; // Base of the numerical system.
TT xx{number}; // Number to convert.
TT exponent{}; // Exponent.
std::stack<TT> fractional_part{}; // Values in the fractional part.
do {
fractional_part.push(xx%kBase);
xx /= kBase;
exponent++;
} while (xx > kBase);
std::cout << xx << '.';
while (!fractional_part.empty()) {
std::cout << fractional_part.top();
fractional_part.pop();
}
std::cout << " x 10^" << exponent << std::endl;
}
int main () {
uint8_t number_1{255};
PrintNumberScientificNotation(number_1); // Does not work.
uint16_t number_2{255};
PrintNumberScientificNotation(number_2); // Works.
uint16_t number_3{65'535};
PrintNumberScientificNotation(number_3); // Works.
uint32_t number_4{4'294'967'295};
PrintNumberScientificNotation(number_4); // Works.
uint64_t number_5{18'446'744'073'709'551'615};
PrintNumberScientificNotation(number_5); // Works.
}
输出:
. x 10^
2.55 x 10^2
6.5535 x 10^4
4.294967295 x 10^9
1.8446744073709551615 x 10^19
据我了解,uint8_t
可以表示无符号整数,最大为255(包括255)(UINT8_MAX
)。为什么我可以代表所有其他表示形式的最大值?
答案 0 :(得分:4)
您的代码的数学部分很好,打印损坏了。
如果您将cout
用于uint_t
,它将把uint_t
解释为字符代码。这是因为uint_t
是无符号字符的类型别名。
可能的解决方法是将其显式转换为整数:
std::cout << unsigned(xx) << '.';
while (!fractional_part.empty()) {
std::cout << unsigned(fractional_part.top());
fractional_part.pop();
}
std::cout << " x 10^" << unsigned(exponent) << std::endl;
答案 1 :(得分:3)
uint8_t被视为一个字符,因此它将打印带有ASCII码的字符。在打印之前,必须将uint8_t值转换为unsigned,如下所示:
uint8_t number_1{255};
PrintNumberScientificNotation(unsigned(number_1));