make fold.expresion.cpp work with negative values

[Delta456]

resolve #267

Description

Please describe the motivation of this pull request, and what problem was solved in this PR.

Change List

• Resolve the error content of Chapter 2 exercise answer bug #267

Reference

I added a static_cast<double> to make it work with negative values.

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说明

此处详细说明 PR 的动机是什么、解决了什么样的问题。

变化箱单

• 修复了 XXX 的 typo 错误
• 增加了 XXX 相关的说明
• 解决了关于 XXX 的描述性错误

参考文献

如果有请注明

[BaiLei27]

Do not use 'auto' for the return of template functions, but use 'double' instead

[frederick-vs-ja]

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);

[Devanshu-0808]

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]

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]

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; } }