关于模板函数中向量迭代器的问题

2022-01-10 00:00:00 iterator templates vector c++ stl

我正在尝试学习 STL 库，但遇到了一个奇怪的问题.这段代码编译完美:

I'm trying to learn the STL library and I'm having a weird problem. This code compiles perfectly:

void Show(vector<int> myvec)
{
    vector<int>::iterator it;
    cout << "Vector contains:";
    for( it = myvec.begin(); it < myvec.end(); it++) 
    {
         cout << " " << *it;
    }
    cout << endl;
}

虽然这个在编译时给了我一条错误消息:

while this one gives me an error message at compile time:

template <class T> 
void Show2(vector<T> myvec)
{
    vector<T>::iterator it;
    cout << "Vector contains:";
    for( it = myvec.begin(); it < myvec.end(); it++)
    {
         cout << " " << *it;
    }
    cout << endl;
}

错误是:

$ g++ hello.cpp
hello.cpp: In function ‘void Show2(std::vector<T, std::allocator<_Tp1> >)’:
hello.cpp:19: error: expected ‘;’ before ‘it’
hello.cpp:21: error: ‘it’ was not declared in this scope

这似乎是一个非常简单的错误，但我找不到它.

It seems a very simple mistake, but I couldn't find it.

推荐答案

你需要说typename vector::iterator it.

另一方面，您正在按值传递 vector.这意味着整个 vector 在函数调用中被复制.void Show(vector<T> const &myvec) 并使用 const_iterator 会更明智.

On another note, you're passing vectors by value. That means the entire vector gets copied in the function call. void Show(vector<T> const &myvec) and using const_iterator would be wiser.

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