Question

解决方案近一年来，我认为我完全理解指针，现在它失败了。如果需要，我会发布整个文件。

// 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;
};

Update功能是问题所在。我的意图是删除（或摆脱）原始。用new You()替换它并完成它。

Answer 1

您将You的指针值传递给void Update(You*,int)。因此update = youTwo;这对currentYou没有影响。

将Update更改为void Update(You*& update, in n) { //...并阅读参考资料。

是的，你有内存泄漏。您更新指针，但永远不会释放旧的currentYou和新的currentYou。你应该使用＆＃34;智能指针＆＃34; （明确地shared_ptr<You>）清理你身后的一切，而不必每次都打电话给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;
};

内存泄漏是固定的（添加了delete），指针现在是指针ref。效果很好。

传递给函数时的C ++指针问题

2 个答案: