函数调用值丢失后 C 返回结构指针

Question

我试图找出为什么我的结构值在函数返回后丢失。

struct A {
   int val1;
   int val2;
}

main() {
   struct A *a;
   a->val1 = 1;
   a->val2 = 2;
   calc((void*)a);
   // calc returns and a's values are 1 and 2 respectively
}

void calc(void* v) {
     struct A *a = (struct A*) v;
     ...
     a->val1 = 2;
     a->val2 = 3;
}

Answer 1

当你想处理指针时，你必须引入动态内存分配。创建指针意味着您告诉编译器“我将在其中存储变量的地址”。因此，在这种情况下，要么创建一个对象，为其分配地址，要么动态为其分配内存。

案例1:

main() {
   struct A *a;
   struct A obj;//assign address of other object manually
   a = &obj;
   a->val1 = 1;
   a->val2 = 2;
   calc((void*)a);
   // calc returns and a's values are 1 and 2 respectively
}

案例2:

main() {
   struct A *a;
   a = (A *) malloc(sizeof(A));//dynamic memory allocation
   a->val1 = 1;
   a->val2 = 2;
   calc((void*)a);
   // calc returns and a's values are 1 and 2 respectively
}