Questions on compound literals in C

Question

COMPOUND LITERALS:

#include<stdio.h>

struct s{
 int x;
 int y;
};

int main(){
 int j = 5;
 printf("\n &j : %p \n",&j);
 struct s *p = &(struct s){j++};
 printf("\n p : %p &j : %p x : %d y : %d \n",p,&j,p->x,p->y);
 return 0;
}

o/p:
-----
&j : 0x7fff416b74ec 
 p : 0x7fff416b74d0 &j : 0x7fff416b74ec x : 5 y : 0

a] Why is p not holding the address of j ?

b] why is j not being typecast to struct s ?

Answer 1

The statement

struct s *p = &(struct s){j++}; // compiler should raise a warning 

is incomplete. It should be like

struct s *p = &(struct s){.x = j++};  

or

struct s *p = &(struct s){.y = j++};  

The uninitialized member will be set ti 0 by default .

The reason that p doesn't hold the address of j is clear that p is holding the address of new object (compound literal), not the address of j itself.

Answer 2

a) p not holds the address of j because p point to freshly created structure. if you want, that p point to j, you should write:

int * p = &j;

b) j is int local (stack) variable, so i see no way, how it can be cast into the structure.

@David C. Rankin, i've reformed my answer

Answer 3

for your first question,, p is of struct type and j is of integer type and you have to make the pointer and the variable of same datatype... like int *p = &j; //where j is of int type.

Answer 4

j是整数类型（原始数据），并且您尝试将其转换为复杂数据类型（对象）..因此，您正在尝试进行装箱和拆箱..这不是c或c ++的概念...