有没有什么方法访问C ++中外部范围的局部变量？ [英] Is there any way to access a local variable in outer scope in C++?

问题描述

只是出于好奇心：如果我有嵌套的范围，像这个示例C ++代码

  using namespace std; 
 
 int v = 1; // global 
 
 int main（void）
 {
 int v = 2; // local 
 {
 int v = 3; // within subscope 
 cout<< subscope：< v<< endl; 
 // cout<< local：< v<< endl; 
 cout<< global：< :: v< endl; 
} 
 cout<< local：< v<< endl; 
 
 cout<< global：< :: v< endl; 
 
} 
  

有任何方法访问变量 v 与中间作用域（既不是全局也不是局部）的值 2 ？

 您可以将新引用声明为别名， int main（void）
 {
 int v = 2; // local 
 int& vlocal = v; 
 {
 int v = 3; // within subscope 
 cout<< local：< vlocal< endl; 
} 
} 
  

但我会避免这种做法。我花了几个小时调试这样一个结构，因为一个变量显示在调试器中，因为范围的改变，我不知道它是如何改变。

Just out of curiosity: if I have nested scopes, like in this sample C++ code

using namespace std;

int v = 1; // global

int main (void)
{
    int v = 2; // local
    {
        int v = 3; // within subscope
        cout << "subscope: " << v << endl;
        // cout << "local: " << v << endl; 
        cout << "global: " << ::v << endl;
    }
    cout << "local: " << v << endl;

    cout << "global: " << ::v << endl;

}

Is there any way to access the variable v with the value 2 from the "intermediate" scope (neither global nor local)?

解决方案

You can declare a new reference as an alias like so

int main (void)
{
    int v = 2; // local 
    int &vlocal = v;
    {
        int v = 3; // within subscope
        cout << "local: " << vlocal  << endl; 
    }
}

But I would avoid this practice this altogether. I have spent hours debugging such a construct because a variable was displayed in debugger as changed because of scope and I couldn't figure out how it got changed.

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