在C ++中，当我将值传递给函数时，是否总是将其转换为适当的类型? [英] In C++, when I pass a value into a function, is it always converted to appropiate type?

问题描述

如果我有一个像void func(size_t x)这样的函数并且调用了函数func(5)，是否立即将5转换为size_t类型?通常适用于所有类型吗?

If I have a function like void func(size_t x) and I call the function func(5), is 5 immediately converted to size_t type? Does this hold generally for all types?

我问是因为我发誓我见过人们在写代码的地方执行func(5.0)(将5作为双精度值传递)或func(0UL)(将0作为无符号long int传递值).这真的有必要吗?我们不能随便传入什么，C ++会将其视为我用来定义函数的类型吗?

I ask because I swear I've seen people write code where they do stuff like func(5.0) (passing 5 as a double) or func(0UL) (passing 0 as an unsigned long int). Is this really necessary? Can't we just pass in whatever we want, and C++ will treat it as the type that I used to define the function?

推荐答案

如果参数类型和传递给函数的类型之间存在隐式转换，则将转换参数.如果没有，例如尝试将std::list传递给需要std::vector的函数，那么它将不会，并且您会得到一个错误.

If there is an implicit conversion between the argument type and the type passed to the function then the argument will be converted. If there isn't one, like trying to pass a std::list to a function that expects a std::vector then it won't and you will get an error.

使用特定文字如func(5.0)或func(5UL)的一个原因是func是模板函数，还是被重载.如果func是模板或过载(对于适当的类型)，则func(5.0)，func(5UL)和func(5)将生成/调用3个不同的函数.一个用于double，一个用于unsigned long，一个用于int.这可能是有意义的，因为可能会有专门化/超负荷对这些类型进行不同的处理.

One reason to use a specific literal, like func(5.0) or func(5UL) is if func is a template function, or is overloaded. If func is a template or is overloaded (for the appropriate types) then func(5.0), func(5UL) and func(5) would generate/call 3 different functions. One for a double, one for a unsigned long and one for an int. This could be meaningful as there could be specializations/overloads handling these types differently.

您还会遇到类似std::accumulate的情况，其第三个参数累加器具有自己的模板类型.假设您想对std::vector<double>中的所有元素求和.如果您使用

You also run into cases like std::accumulate whose third parameter, the accumulator, has its own template type. Lets say you want to sum all of the elements in a std::vector<double>. If you use

std::accumulate(vector.begin(), vector.end(), 0)

那么您得到的结果将不同于

then you would get a different result than

std::accumulate(vector.begin(), vector.end(), 0.0)

因为第一次调用使用int存储每次将截断的总和，而后者使用double则您将获得预期的结果.

because the first call uses an int to store the sum which will truncate each time, while the latter uses a double and you will get the expected result.

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