关于c ++递归和局部变量的问题 [英] question on c++ recursion and local variables

问题描述

假设我有这个递归：

  void doSomething（double j）
 {
 double x ; 
 double y; 
 
 x = j -1; 
 y = j -2; 
 
 doSomething（x + y）; 
 
 x = j + 31; 
 y = j + 12; 
} 
  

我知道这个递归执行无限，

我的问题是关于在递归树中的变量x和y的范围...将x和y的范围仅对递归树中特定阶段的函数有效吗？或者当我再次调用doSomething（）时，当递归树中的子doSomething（）重命名x和y时，它将重置父类的x和y变量，还是创建一个全新的x和y变量对于递归树中的那个阶段？

解决方案

将x和y的范围仅对递归中特定阶段中的函数有效树？

是的。

doSomething（），并且递归树中的子doSomething（）重定义x和y，它将重置父类的x和y变量

否。

是在递归树中创建一个全新的x和y变量只有？

是的。

strong>
此示例应该是有帮助的。

  #include< iostream> 
 
无效的foo（int temp）
 {
 int num = temp; 
 
 if（temp == 0）
 return; 
 
 foo（temp-1）; 
 
 std :: cout<< &＃< \t<< num<< \\\
; 
} 
 
 int main（）
 {
 foo（5）; 
 return 0; 
} 
  

输出：

0xbfa4e2d0 1

0xbfa4e300 2

0xbfa4e330 3

0xbfa4e360 4

0xbfa4e390 5

注意 num 的地址不同，它自己的值 num 。
Ideone

suppose I have this recursion:

void doSomething(double j)
{
    double x;
    double y;

    x = j -1;
    y = j -2 ;

    doSomething(x+y);

    x = j + 31;
    y = j + 12 ;
}

I know that this recursion executes infinitely, but just ignore that

My question is with regards to variables x and y's scope in the recursion tree...will x and y's scope be valid only for the function in that specific stage in the recursion tree? or when I call doSomething() again, when the child doSomething() in the recursion tree redeclares x and y, will it reset the parents' x and y variables as well or is it creating an entirely new x and y variables that is valid for that stage in the recursion tree only?

解决方案

will x and y's scope be valid only for the function in that specific stage in the recursion tree?

Yes.

when I call doSomething() again, and the child doSomething() in the recursion tree, redeclares x and y, will it reset the parents' x and y variables as well

No.

is it creating an entirely new x and y variables that is valid for that stage in the recursion tree only?

Yes.

Edit 1: This example should be helpful.

#include <iostream>

void foo( int temp )
{
     int num = temp;

     if( temp == 0)
          return; 

     foo(temp-1) ;

     std::cout << &num << "\t" << num << "\n" ;
}

int main()
{
     foo(5) ;
     return 0;
}

Output:

0xbfa4e2d0 1
0xbfa4e300 2
0xbfa4e330 3
0xbfa4e360 4
0xbfa4e390 5

Notice the address of num being different and each call has it's own value of num. Ideone

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