Question

以下是代码：

 // allocation  
void allocateSymbolStorage(char **pepperShakerList, char **pepperList)
{   
    //  allocate storage for an array of pointers
    pepperShakerList = (char **) malloc(MAX_PEPPER_SHAKERS * sizeof(char *));

    for (int i = 0; i < MAX_PEPPER_SHAKERS; i++)
    {
        if ((pepperShakerList[i] = (char *) malloc(MAX_SHAKERNAME_LENGTH * sizeof(char))) == NULL)
            fatalError("failed pepperShakerList alloc");
    }

    //  allocate storage for an array of pointers 
    pepperList = (char **) malloc(MAX_PEPPERS * sizeof(char *));

    for (int i = 0; i < MAX_PEPPERS; i++)
    {
        if ((pepperList[i] = (char *) malloc(MAX_PEPPER_LENGTH * sizeof(char))) == NULL)
            fatalError("failed pepperList alloc");
    }
}

void buildPepperShakers(void)
{
    char **pepperShakerList, **pepperList; 

    allocateSymbolStorage(pepperShakerList, pepperList);

    // ....

    freeSymbolStorage(pepperShakerList, pepperList);
}

这是VS 2010错误：：警告C4700：未初始化的局部变量'pepperList'使用

这是混乱：如果在分配函数中分配char **，为什么会出错？这是不属于范围的问题吗？

Answer 1

这就是你想要的，你需要取消引用传入的指针：

*pepperShakerList = (char *) malloc(MAX_PEPPER_SHAKERS * sizeof(char *));

Answer 2

假设你正在讨论的是pepperList而不是symbolList，并且假设您在allocationSymbolStorage中的代码反映了您想要做的事情，那么VC正在抱怨。

现在你的代码会崩溃，因为在buildPepperShakers（）中你没有从allocateSymbolStorage获得任何值。

因此，您的allocateSymbolStorage应声明为：

void allocateSymbolStorage(char ***pepperShakerList, char ***pepperList)

然后将buildPepperShakers中的本地指针持有者变量的地址，即pepperList和pepperShakerList传递给分配函数，以便它可以根据TJD的答案进行分配。那就是：

void buildPepperShakers(void) {
       char **pepperShakerList, **pepperList;
       allocateSymbolStorage(&pepperShakerList, &pepperList); 
}

当然，您的allocateSymbolStorage正文现在变为：

void allocateSymbolStorage(char ***pepperShakerList_p, char ***pepperList_p)
{   
    char **pepperShakerList, **pepperList;
    //  allocate storage for an array of pointers
    pepperShakerList = (char **) malloc(MAX_PEPPER_SHAKERS * sizeof(char *));

    for (int i = 0; i < MAX_PEPPER_SHAKERS; i++)
    {
        if ((pepperShakerList[i] = (char *) malloc(MAX_SHAKERNAME_LENGTH * sizeof(char))) == NULL)
            fatalError("failed pepperShakerList alloc");
    }

    //  allocate storage for an array of pointers 
    pepperList = (char **) malloc(MAX_PEPPERS * sizeof(char *));

    for (int i = 0; i < MAX_PEPPERS; i++)
    {
        if ((pepperList[i] = (char *) malloc(MAX_PEPPER_LENGTH * sizeof(char))) == NULL)
            fatalError("failed pepperList alloc");
    }

    *pepperShakerList_p = pepperShakerList;
    *pepperList_p = pepperList;

}

现在VC不应该抱怨。虽然这是对对象进行内存管理的一种丑陋方式:-)

malloc，范围，初始化（或缺少）

2 个答案: