释放分配给空指针数组的内存 [英] Freeing memory allocated to an array of void pointers

问题描述

我声明了一个空指针数组.每个都指向一个任意类型的值.
void **values; // Array of void pointers to each value of arbitary type

I am declaring an array of void pointers. Each of which points to a value of arbitary type.
void **values; // Array of void pointers to each value of arbitary type

初始化值如下:

    values = (void**)calloc(3,sizeof(void*));
    //can initialize values as: values = new void* [3];
    int ival = 1;
    float fval = 2.0;
    char* str = "word";
    values[0] = (void*)new int(ival);
    values[1] = (void*)new float(fval);
    values[2] = (void*)str;

    //Trying to Clear the memory allocated
    free(*values); 
    //Error: *** glibc detected *** simpleSQL: free(): invalid pointer: 0x080611b4
    //Core dumped
    delete[] values*;
    //warning: deleting 'void*' is undefined
    //Similar Error.

现在如何释放/删除分配给值的内存(无效指针数组)?

Now how do I free/delete the memory allocated for values ( the array of void pointers)?

推荐答案

您有3种动态分配的东西，需要以2种不同的方式释放它们:

You have 3 things that are dynamically allocated that need to be freed in 2 different ways:

delete reinterpret_cast<int*>( values[0]);    
delete reinterpret_cast<float*>( values[1]);

free( values); // I'm not sure why this would have failed in your example, 
               //    but it would have leaked the 2 items that you allocated 
               //    with new

请注意，由于str并不是动态分配的，因此它(实际上不能)不应被释放.

Note that since str is not dynamically allocated it should not (actually cannot) be freed.

一些注意事项:

• 我假设sizeof(void) 本来是sizeof(void*) 因为您将无法编译
• 我不会说任何关于 您看似随机的演员，除了 看起来已经准备好了的代码 普遍的灾难

• I'm assuming that the sizeof(void) was meant to be sizeof(void*) since what you have won't compile
• I'm not going to say anything about your seemingly random casting except that it looks like code that ready for disaster in general

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