通过引用传递动态var的指针 [英] Passing pointer of dynamic var by reference

问题描述

我试图创建动态变量并通过引用在 new_test 函数中传递它的地址，但它不工作。

 
 code> #include< iostream> 
 using namespace std; 
 
 struct test 
 {
 int a; 
 int b; 
}; 
 
 void new_test（test * ptr，int a，int b）
 {
 ptr = new test; 
 ptr  - > a = a; 
 ptr  - > b = b; 
 cout<< ptr：<< ptr<< endl; //这里显示内存地址
}; 
 
 int main（）
 {
 
测试* test1 = NULL; 
 
 new_test（test1，2，4）; 
 
 cout<< test1：<< test1<< endl; // always 0  -  why？ 
 delete test1; 
 
 return 0; 
} 
  

解决方案

代码不通过指针，因此对参数 ptr 的更改对于函数是本地的，对调用者不可见。更改为：

  void new_test（test *& ptr，int a，int b）
 // ^ b $ b  

I'm trying to create dynamic variable and pass its address by reference within new_test function, but it doesn't work. What am I doing wrong?

The code:

#include <iostream>
using namespace std;

struct test
{   
    int a;
    int b;
};  

void new_test(test *ptr, int a, int b)
{   
    ptr = new test;
    ptr -> a = a;
    ptr -> b = b;
    cout << "ptr:   " << ptr << endl; // here displays memory address
};  

int main()
{   

    test *test1 = NULL;

    new_test(test1, 2, 4); 

    cout << "test1: " << test1 << endl; // always 0 - why?
    delete test1;

    return 0;
}

解决方案

The code does not pass the pointer by reference so changes to the parameter ptr are local to the function and not visible to the caller. Change to:

void new_test (test*& ptr, int a, int b)
                  //^

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