make fold.expresion.cpp work with negative values
#268
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Do not use 'auto' for the return of template functions, but use 'double' instead |
| @@ -11,8 +11,8 @@ | |||
| #include <iostream> | |||
| template<typename ... T> | |||
| auto average(T ... t) { | |||
| return (t + ... ) / sizeof...(t); | |||
| return static_cast<double>((t + ... )) / sizeof...(t); | |||
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我个人建议是模仿 <cmath> 中浮点函数的额外重载，把整数类型转换为 double 而浮点类型不变。这样可以避免 long double 被转换为 double。
Suggested change
| return static_cast<double>((t + ... )) / sizeof...(t); | |
| constexpr auto integer_to_double = [](auto x) { | |
| if constexpr (std::is_integral_v<decltype(x)>) | |
| return static_cast<double>(x); | |
| else | |
| return x; | |
| }; | |
| return (integer_to_double(t) + ... ) / sizeof...(t); |
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static_caste() it will explicitly change the data type of t to double
#include <iostream>
template<typename ... T>
auto average(T ... t) {
return (static_cast<double>(t) + ...) / sizeof...(t);
}
int main() {
std::cout << average(1, 2, 3, 4, 5, -6, -7, -8, -9, -10) << std::endl;
}
}
|
static_caste() it will explicitly change the data type of t to double #include <iostream>
template<typename ... T>
auto average(T ... t) {
return (static_cast<double>(t) + ...) / sizeof...(t);
}
int main() {
std::cout << average(1, 2, 3, 4, 5, -6, -7, -8, -9, -10) << std::endl;
} |
Devanshu-0808
suggested changes
Jun 8, 2024
| @@ -11,8 +11,8 @@ | |||
| #include <iostream> | |||
| template<typename ... T> | |||
| auto average(T ... t) { | |||
| return (t + ... ) / sizeof...(t); | |||
| return static_cast<double>((t + ... )) / sizeof...(t); | |||
There was a problem hiding this comment.
static_caste() it will explicitly change the data type of t to double
#include <iostream>
template<typename ... T>
auto average(T ... t) {
return (static_cast<double>(t) + ...) / sizeof...(t);
}
int main() {
std::cout << average(1, 2, 3, 4, 5, -6, -7, -8, -9, -10) << std::endl;
}
}
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resolve #267
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I added a
static_cast<double>to make it work with negative values.说明
此处详细说明 PR 的动机是什么、解决了什么样的问题。
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