带有默认值的模板返回类型 [英] Template Return Type with Default Value

问题描述

所以在编写一个C ++模板类时，我定义了一个返回模板类型对象的方法，如下所示：

So in writing a C++ template class, I have defined a method that returns an object of the templated type, as such:

template <typename T>
class Foo
{
public:
    T GetFoo()
    {
        T value;

        //Do some stuff that might or might not set the value of 'value'

        return value;
    }
 };

int main() {

    Foo<int> foo;

    foo.GetFoo();

    return 0;
}

这会产生以下警告：

prog.cpp: In member function ‘T Foo<T>::GetFoo() [with T = int]’:
prog.cpp:15: warning: ‘value’ is used uninitialized in this function

我理解为什么会发生这种情况 - 作为 GetFoo 一部分的未初始化 int 。事情是，如果我使用 Foo< SomeClass> ，行 T值; 将初始化 value 使用 SomeClass 的默认构造函数。

I understand why this is happening - I am returning an uninitialized int as part of GetFoo. The thing is, if I were to use Foo<SomeClass>, the line T value; would initialize value using the default constructor of SomeClass.

通过执行以下操作抑制此警告：

I have managed to suppress this warning by doing the following:

    T GetFoo()
    {
        T value = T();

        //Do some stuff that might or might not set the value of 'value'

        return value;
    }

这似乎适用于原始类型（例如 int 和 float ）和类，至少只要该类有一个默认的构造函数和拷贝构造函数。我的问题是 - 这是接受的解决这个问题的方法吗？是否有任何副作用我应该知道吗？

This seems to work for primitive types (such as int and float) and classes, at least so long as that class has a default constructor and copy constructor. My question is - is this the accepted way of solving this problem? Are there any side effects of this I should know about?

推荐答案

听起来OK，如果类没有复制构造函数，无法返回。

Sounds OK, if the class has no copy constructor, you will not be able to return it anyway.

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