C++ 错误：没有忽略无效值，因为它应该是
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error: void value not ignored as it ought to be
提问by Aquarius_Girl
template <typename Z> Z myTemplate <Z> :: popFromVector ()
{
if (myVector.empty () == false)
return myVector.pop_back ();
return 0;
}
int main ()
{
myTemplate <int> obj;
std :: cout << obj.popFromVector();
return 0;
}
Error:
错误：
error: void value not ignored as it ought to be
AFAI can see, the return type of popFromVectoris NOT void. What's the point that I am missing?
The error disappears when I comment out this call in main().
AFAI 可以看到，返回类型popFromVector不是 void。我失踪的重点是什么？当我在 main() 中注释掉这个调用时，错误消失了。
采纳答案by Puppy
std::vector<T>::pop_back()returns void. You attempt to return it as an int. This is not allowed.
std::vector<T>::pop_back()返回无效。您尝试将其作为 int 返回。这是不允许的。
回答by Jason
That is because the definition of std::vector::pop_backhas a voidreturn type ... you are trying to return something from that method, which won't work since that method doesn't return anything.
那是因为 的定义std::vector::pop_back有一个void返回类型......你试图从那个方法返回一些东西，这将不起作用，因为该方法不返回任何东西。
Change your function to the following so you can return what's there, and remove the back of the vector as well:
将您的函数更改为以下内容，以便您可以返回那里的内容，并删除向量的背面：
template <typename Z> Z myTemplate <Z> :: popFromVector ()
{
//create a default Z-type object ... this should be a value you can easily
//recognize as a default null-type, such as 0, etc. depending on the type
Z temp = Z();
if (myVector.empty () == false)
{
temp = myVector.back();
myVector.pop_back();
return temp;
}
//don't return 0 since you can end-up with a template that
//has a non-integral type that won't work for the template return type
return temp;
}
回答by ltjax
It's the pop_back(). It has a void return type. You have to use back()to get the actual value. This is to avoid unnecessary copies.
这是pop_back(). 它有一个 void 返回类型。您必须使用back()来获取实际值。这是为了避免不必要的副本。
