C ++释放了struct使用的所有内存 [英] C++ free all memory used by struct

问题描述

快速提问；我已经四处搜寻，并且已经找到了一些答案，但是我有点偏执，所以我想确定.

Quick question; I've googled around and found some answers already, but I'm a bit paranoid so I want to be sure.

考虑这种情况:

struct CoordLocation
{
    float X;
    float Y;
    float Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

调用delete是否还会清除X，Y，Z字段使用的内存?我发现一些答案提到我只是删除POINTER，而不是这样删除实际引用的对象. 如果...

Will calling delete also clear the memory used by the fields X, Y, Z? Some answers I found mentioned that I'd just delete the POINTER, not the actually referenced object this way. What if...

struct CoordLocation
{
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

如果我为结构的构造函数/析构函数中的每个对象手动释放内存该怎么办?

And what if I manually free the memory for each object inside the struct's constructor/destructor?

struct CoordLocation
{
    CoordLocation()
    {
         *X = new float;
         *Y = new float;
         *Z = new float;
    }
    ~CoordLocation()
    {
         delete X; delete Y; delete Z;
    }
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

我注意到这种情况很简单，例如:

I noticed that for a simple situation such as:

   float *a = new float;
   *a = 5.0f;
   printf("%f", *a);
   delete a;
   printf("%f", &a);

printf会打印5.0，因此a所指向的变量不会被完全破坏.

printf would print 5.0, so the variable pointed to by a is not exactly destroyed.

所以我的问题是: 在这种情况下，如何可靠地释放(如无内存泄漏)该结构使用的所有内存?

So my question is: How can I reliably free (as in no memory leaks) ALL the memory used by the struct in this case?

struct CoordLocation
{
    float X;
    float Y;
    float Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

谢谢！

推荐答案

您只需要delete用new分配的内存.

You only need to delete memory you allocate with new.

printf会打印5.0，因此a所指向的变量不会被完全破坏.

printf would print 5.0, so the variable pointed to by a is not exactly destroyed.

您实际上正在遇到未定义的行为.尽管该值仍然存在，但是内存已释放并且可以重复使用.

You're actually running into undefined behavior. Although the value is still there, the memory was released and can be reused.

以下内容:

struct CoordLocation
{
    float X;
    float Y;
    float Z;
};

如果省略析构函数，则不会造成内存泄漏.

can't create a memory leak if you omit the destructor.

您的下一个片段:

struct CoordLocation
{
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

可能会造成内存泄漏，但并非如此.将执行以下操作:

can potentially create a memory leak, but not as it is. The following will:

int main()
{
    CoordLocation *coord = new CoordLocation();
    coord->X = new float();
    delete coord;
    return 0;
}

您的第三个示例

struct CoordLocation
{
    CoordLocation()
    {
         *X = new float;
         *Y = new float;
         *Z = new float;
    }
    ~CoordLocation()
    {
         delete X; delete Y; delete Z;
    }
    float *X;
    float *Y;
    float *Z;
};

int main()
{
    CoordLocation *coord = new CoordLocation();
    delete coord;
    return 0;
}

不会造成内存泄漏，因为您释放了分配的所有内存.如果您要省略析构函数或忘记调用delete coord;，则它们将发生内存泄漏.

won't create a memory leak because you free all the memory that you allocate. If you were to omit the destructor or forget to call delete coord;, they you'd have a memory leak.

一个好的经验法则:每个new都调用一个delete，每个new[]都调用一个delete[]，这很安全.

A good rule of thumb: call a delete for every new and a delete[] for every new[] and you're safe.

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