c++ - 变量作用域可不可以扩展

Question

fun(int &a)  
{
         
}


main()   
{     
    ........     
    int a = 5 ;    
    fun(a);    
}

调用函数fun，传变量a的引用，在main中使用a，和在fun中使用a的效果是一样的
可不可以理解为变量a的作用域从main扩展到fun

Answer 1

The concept of scope is the scope of declaration. It has nothing to do with the instance.

Answer 2

No. Simply change the parameter name to fun(int &b), can the scope of variable a be extended to fun

Answer 3

Purely guessworkWhy does C++ choose & as the reference symbol? I think reference is actually a kind of address, which can be compared to a pointer. A space will be opened in the fun function stack to store the address of variable a. To access a in fun, you may access a indirectly through the address

Answer 4

There is no variable scope expansion, it is just a reference transfer of variables

Answer 5

This question is so retarded, I want to ask you how you learned it