Structures and Pointers dereferencing question-CodePudding

The pointer MyStruct* ps points to &s. What is the value, if any, after de-referencing only *ps without accessing any struct members of MyStruct?

#include    <stdio.h>

typedef struct {
    char c;
    int x;
    double d;
} MyStruct;

int main(void) {
    MyStruct s = {'a', 132, 456.789};
    MyStruct* ps = &s;
    
    printf("Member c has a value of %c\n",(*ps).c);
    printf("Member x has a value of %d\n", (*ps).x);
    printf("Member d has a value of %f\n", (*ps).d);    

    /* printf("Here i want to print the outcome of\n", *ps); */
}

EDIT: I finally got the solution! And i know what its containing as value. It's the char 'a'. With printf("%c\n", *ps i got the value inside of *ps. I know i get a warning but still i know now what it contains. Also the comments and other answers were really helpful!

CodePudding user response：

The definition

MyStruct s = {'a', 132, 456.789};

creates a block in memory looking something like

 --- 
| c |
 --- 
| x |
 --- 
| d |
 ---

Then the definition

MyStruct* ps = &s;

modifies it to something like this:

 ----       --- 
| ps | --> | c |
 ----       --- 
           | x |
            --- 
           | d |
            ---

The "value" of ps is the address of the first member of the s structure object.

CodePudding user response：

*ps denotes the entire structure. The “value” of *ps is an aggregate with the values of all its members. The C standard does not provide any way to print it, but you can assign it to another structure of the same type, pass it to a function that accepts an argument of that type, and return it from a function that has a return type of that type. Effectively, *ps is the same as s.

CodePudding user response：

one way to think of it is that

 MyStruct* ps;

is the same as

 MyStruct *ps;

Which says: '*ps' is a 'MyStruct'