如何将typename作为函数中的参数？ （C ++）

I need to be able to pass a typename as a parameter:

我需要能够将typename作为参数传递:

int X = FileRead(file, 9, char);

The concept is for FileRead(std::fstream, int pos, ???) to read pos*sizeof(whatever the type is) to get the desired position. I tried templates:

这个概念是为FileRead(std :: fstream,int pos,???)读取pos * sizeof(无论类型是什么)来获得所需的位置。我试过模板:

template<typename T>
T FileRead(std::fstream file, int pos, T type)
{
    T data;
    file.read(reinterpret_cast<char*>(&data), sizeof(data));
    return data;
}

but that required that I create a variable of the type to use every time I wanted to use FileRead, and I really don't feel like redesigning an entire program just because of one function, so is there anyway to use a typename as a parameter?

但这需要我创建一个每次我想使用FileRead时使用的类型的变量,我真的不想因为一个函数而重新设计整个程序,所以无论如何都要使用typename作为参数?

3 个解决方案

#1

To use the name of a type as a parameter, use a template.

要使用类型名称作为参数,请使用模板。

template<typename T>
T FileRead(std::fstream &file, int pos)
{
    T data;
    file.read(reinterpret_cast<char*>(&data), sizeof(T));
    return data;
}

This assumes that the type is default constructible. If it is not, I guess you would have difficulty streaming it out of a file anyway.

这假定该类型是默认可构造的。如果不是,我想你无论如何都难以将它从文件中流出来。

Call it like this:

这样叫:

char value=FileRead<char>(file, pos);

If you do not want to have to specify the type in the call, you could modify your API:

如果您不想在调用中指定类型,则可以修改API:

template<typename T>
void FileRead(std::fstream &file, int pos, T &data)
{
    file.read(reinterpret_cast<char*>(&data), sizeof(T));
}

Then call it like this - the type is inferred:

然后像这样调用它 - 推断类型:

char value;
FileRead(file, pos, value);

#2

Very simple:

template<typename T>
T FileRead(std::fstream file, int pos)
{
    T data;
    file.read(reinterpret_cast<char*>(&data), sizeof(data));
    return data;
}

and call it via:

并通过以下方式调用:

char x = FileRead<char>(file, pos);

#3

There is no such things as types once your program is compiled. This is the style of C++.

编译程序后,没有类型的东西。这是C ++的风格。