Question

我很长时间以来一直在摸不着头脑。

typedef struct B_{
   /* something */
}B;

typedef struct A_{
   B* pointB;
}A;

func1(A *pointA)
{
   pointA->pointB = malloc(SOMESIZE);
}

function(A *pointA)
{
   /* STATE 1 
    * pointA is perfectly allocated here and 
    * pointB inside pointA is NULL at this moment.
    */

   func1(pointA);

   /* STATE 2
    * pointB is still NULL at this moment??
    */
}

在function()中，参数pointA是完全分配的结构，具有多个成员和指针pointB，即NULL。我正在调用另一个以pointA为参数并分配pointB的函数。我调试了pointB malloc之后的func1()，但是当我从pointB返回时，我仍然看到NULL为pointA。我认为这看起来像我传递{{1}}作为价值而不是参考，但我很难将这一点放到我的头脑中。我传递了一个完全有效的结构指针，其成员指针为NULL，并且我分配了该成员指针。

Answer 1

我刚刚尝试了以下程序，请告诉我你的是否类似：

typedef struct B_{
    /* something */   
}B;

typedef struct A_{
    B* pointB; 
}A;

func1(A *pointA)
{
    pointA->pointB = malloc(SOMESIZE);
    printf("\nValue of pointB:%p\n", (void*)(pointA->pointB));
}

function(A *pointA)  // Check in Debugger, For me this is coming as NULL
{
    /* STATE 1 
     * pointA is perfectly allocated here and 
     * pointB inside pointA is NULL at this moment.
    */

    pointA = (A*)malloc(sizeof(A)); 
    if (pointA == NULL) {
       printf("\nReturning NULL from malloc\n");
       exit(0);
    } 
    printf("\nValue of pointA:%p\n", (void*)pointA);
    func1(pointA);
    printf("\nValue of pointB:%p\n", (void*)(pointA->pointB));

    /* STATE 2
     * pointB is still NULL at this moment??
    */
}

int main() 
{
     A *pointA = NULL;

     // Passing pointA as value from here, so it will be NULL after this call.
     function1(pointA);

     return 0;
}

这里，pointA作为值传递给function1（），因此从这个函数返回后，它仍然是NULL。所以，我认为调试器存在一些问题。

pointA的值：0xa39010

pointB的值：0xa39030

malloc之后没有返回结构指针内的指针

1 个答案: