为什么不实例化模板类成员？ [英] Why uncalled template class members aren't instantiated?

问题描述

我想知道当我创建一个类模板的实例并指定模板类型参数。

1）为什么不调用函数不被调用？ 。

2）不会编译，直到我尝试使用它？

3）什么是逻辑背后这种行为？

I was wondering that when I create an instance of a class template with specifying the template type parameter.
1) why the non-called function do not get instatiated ? .
2) dont they get compiled until I try to use it ?
3) what is the logic behind this behavior ?

示例

template <class T>
class cat{
public:

T a;
void show(){
   cout << a[0];
}
void hello(){
   cout << "hello() get called \n";
}
}; 

int main(){
cat<int> ob1; // I know that show() did not get instatiated, otherwise I will get an    error since a is an int
ob1.hello();

 }

推荐答案

为什么？因为在C ++中，它是（在某些情况下，完全不可能，就我所知）说，只有编译这个代码，如果X，Y和Z是真的。

Why? Because in C++, it's difficult (and in some cases, outright impossible, as far as I know) to say, "only compile this code if X, Y, and Z are true".

例如，你怎么说，只有我的复制构造函数如果嵌入对象可以复制？据我所知，你不能。

For example, how would you say, "only my copy constructor if the embedded object can be copied"? As far as I know, you can't.

所以他们只是使他们不编译，除非你实际调用他们。

So they just made them not compile unless you actually call them.

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