傳遞給函數時的C ++指針問題

Question

下面的解決方案近一年來，我以為我完全理解了指針，但是現在失敗了。 如果需要，我將發布整個文件。

// Test Structure and Function
struct You {
    int x;
    int y;
    string str;
};

bool Show(You* showValue);


// Should (delete) in whatever way possible and update its address to the     "You* update" you sent
void Update(You* update, int n) {

    // Create a new "You"
    You* youTwo = new You();
    youTwo->x = 55;
    youTwo->y = 43;
    youTwo->str = "Twin";

    // Update?
    update = youTwo;

    return; 
};




bool Show(You* showValue) {
    cout << "Show:" << endl;
    cout << showValue->x << '\t';
    cout << showValue->y << '\t';
    cout << showValue->str << '\t'; 
    cout << endl << endl;
};



int main(int argc, char** argv) {

    // Original You
    You* currentYou = new You();
    currentYou->x = 1;
    currentYou->y = 2;
    currentYou->str = "You";

    // Update the current you to a new you
    Show(currentYou);   // works
    Update(currentYou, 5);  // no compile errors
    Show(currentYou); // shows initial values instead of the updated

    return 0;
};

問題所在是Update功能。 我的意圖是刪除（或刪除）原始文件。 用new You()替換它並完成它。

Answer 1

你通過指針You按值void Update(You*,int) 因此update = youTwo; 這對currentYou沒有影響。

更改Update到void Update(You*& update, in n) { //...和引用閱讀起來。

順便說一句，您有內存泄漏。 您更新了指針，但從未取消分配舊的currentYou或新的currentYou 。 您應該使用“智能指針”（明確地為shared_ptr<You> ）來清理背后的所有內容，而不必每次都調用delete 。

Answer 2

#include <iostream>
#include <string>
using namespace std;



// Test Structure and Function
    struct You {
    int x;
    int y;
    string str;
};


bool Show(You* showValue);



// Should (delete) in whatever way possible and update its address to the "You* update" you sent
void Update(You*& update, int n) {

// Clean Up
delete update;

// Create a new "You"
You* youTwo = new You();
youTwo->x = 55;
youTwo->y = 43;
youTwo->str = "Twin";

// Update address
update = youTwo;

return; 
};




bool Show(You* showValue) {
cout << "Show:" << endl;
cout << showValue->x << '\t';
cout << showValue->y << '\t';
cout << showValue->str << '\t'; 
cout << endl << endl;
};



int main(int argc, char** argv) {

// Original You
You* currentYou = new You();
currentYou->x = 1;
currentYou->y = 2;
currentYou->str = "You";

// Update the current you to a new you
Show(currentYou);
Update(currentYou, 5); 
Show(currentYou); 

delete currentYou;
return 0;
};

內存泄漏已修復（已添加delete ），並且指針現在是一個指針引用。 完美的作品。