Hi Please explain this concept related to structure

[sathishc58]

Hi All

I read in book that mentioning the structure name gives the whole structure itself and not its base address? What is meant by that?

I wrote a small program like this

struct a
{
int a;
int b;
} *obj, obj1;

main()
{
obj=&obj1;
printf(”Address is %pn”, obj);
}

Output

Address is 80608bc

I guess what I got is base address of the structure. But I am not sure.

We would also read that if we know the address of one member of the structure we can get the base address of the structure. I am really confused by this.

Explanation please…

Thanks & Regards
Sathish Kumar

[ruskalym]

obj is a structure pointer (struct a *) and obj1 is a plain struct (struct a).

When passing obj1 as argument to a function, the structure is entirely duplicated in stack and changes made to it will not be saved in calling function.

Passing the pointer is a gain of memory and time, and permits direct modification of pointed structure.

[ruskalym]

This code :

#include <stdio.h>
 
struct a
{
   int a;
   int b;
} *obj, obj1;
 
void change_struct(struct a mystruct)
{
   mystruct.a = 3;
   mystruct.b = 4;
}
 
void change_struct_pointer(struct a *pmystruct)
{
   pmystruct->a = 5;
   pmystruct->b = 6;
}
 
int main()
{
   obj1.a = 1;
   obj1.b = 2;
   printf("a = %d, b = %d\n", obj1.a, obj1.b);
 
   change_struct(obj1);
   printf("a = %d, b = %d\n", obj1.a, obj1.b);
 
   obj = &obj1;
   change_struct_pointer(obj);
   printf("a = %d, b = %d\n", obj1.a, obj1.b);
 
   return 0;
}

produces following output :

a = 1, b = 2
a = 1, b = 2
a = 5, b = 6

[sathishc58]

Hi


What you have said is correct? But I am sorry. I think your answer is not related to my question.


Thanks & Regards
Sathish Kumar

obj is a structure pointer (struct a *) and obj1 is a plain struct (struct a).

When passing obj1 as argument to a function, the structure is entirely duplicated in stack and changes made to it will not be saved in calling function.

Passing the pointer is a gain of memory and time, and permits direct modification of pointed structure.

[sathishc58]

Hi

Sorry actually I meant to say that your answer is not sufficient for my 2nd point. i.e. If we know the address of one member of structure we can get the base of the address of the structure. Please explain this concept.

Actually I have come across the macro which is used to find the base address of the structure. If you explain it will be very useful

Thanks & Regards
Sathish Kumar

#include <stdio.h>
 
struct a
{
   int a;
   int b;
} *obj, obj1;
 
void change_struct(struct a mystruct)
{
   mystruct.a = 3;
   mystruct.b = 4;
}
 
void change_struct_pointer(struct a *pmystruct)
{
   pmystruct->a = 5;
   pmystruct->b = 6;
}
 
int main()
{
   obj1.a = 1;
   obj1.b = 2;
   printf("a = %d, b = %d\n", obj1.a, obj1.b);
 
   change_struct(obj1);
   printf("a = %d, b = %d\n", obj1.a, obj1.b);
 
   obj = &obj1;
   change_struct_pointer(obj);
   printf("a = %d, b = %d\n", obj1.a, obj1.b);
 
   return 0;
}

produces following output :

a = 1, b = 2
a = 1, b = 2
a = 5, b = 6

[/quote]

[weaknessforcats]

struct a
{
int a;
int b;
} *obj, obj1;

main()
{
obj=&obj1;
printf(”Address is %pn”, obj);
}

Output

Address is 80608bc

I guess what I got is base address of the structure. But I am not sure.

We would also read that if we know the address of one member of the structure we can get the base address of the structure. I am really confused by this.

Explanation please…

OK here goes:

When a struct variable is created, the members are aligned. That is, the compiler sets the position of the various members along word boundaries. A word, in this case, is the processor register size. So, a 32-but processor has a 32-bit word. That means, it is convenient if the variables are set on 32-bit boundaries. Usually, the compiler sees this as the sizeof(int).

This struct:

struct Test
{
    int a:
    char b;
    int c;
};
 

will have the int a on a 32-bit boundary, char b on a 32-bit boudary and int c on a 32-bit boundary. The char is only one byte. Therefore, 3 slack bytes (sometimes call pad bytes) will be added after char b.

Therefore, the sizeof(Test) is not the same as sizeof(Test.a) + sizeof(Test.b) + sizeof(Test.c).

Therefore, if you have the address of a member variable it may not be as easy as you think to deduce the address of the struct variable itself. This is further complicated in C++ when hidden addresses are tuck away in the struct (or class) variable to support object-oriented programming.

[ruskalym]

If you know the address of one of structure's members and know which member it is, you can find out base address by substracting sizes of previous members. This works for packed structure, otherwise you have to take care about memory alignment issues.

[sathishc58]

Hi

Thanks for your explanation. I clearly understood the concept.


1) But I dont understand why we should have such a approach?

2) Can you give me an example why we should go for this or under which circumstance we will need this?

Because I am able to get the base address of the structure by writing this simple code

struct a
{
int a;
int b;
} *obj, obj1;

main()
{
obj=&obj1;
printf(”Address is %pn”, obj);
}

Output

Address is 80608bc


Please explain....


Thanks & Regards
Sathish Kumar

OK here goes:

When a struct variable is created, the members are aligned. That is, the compiler sets the position of the various members along word boundaries. A word, in this case, is the processor register size. So, a 32-but processor has a 32-bit word. That means, it is convenient if the variables are set on 32-bit boundaries. Usually, the compiler sees this as the sizeof(int).

This struct:

struct Test
{
    int a:
    char b;
    int c;
};
 

will have the int a on a 32-bit boundary, char b on a 32-bit boudary and int c on a 32-bit boundary. The char is only one byte. Therefore, 3 slack bytes (sometimes call pad bytes) will be added after char b.

Therefore, the sizeof(Test) is not the same as sizeof(Test.a) + sizeof(Test.b) + sizeof(Test.c).

Therefore, if you have the address of a member variable it may not be as easy as you think to deduce the address of the struct variable itself. This is further complicated in C++ when hidden addresses are tuck away in the struct (or class) variable to support object-oriented programming.

[weaknessforcats]

1) But I dont understand why we should have such a approach?

2) Can you give me an example why we should go for this or under which circumstance we will need this?

Because I am able to get the base address of the structure by writing this simple code

struct a
{
int a;
int b;
} *obj, obj1;

main()
{
obj=&obj1;
printf(”Address is %pn”, obj);
}

Alignment is used to speed up access to the variables. If a variable si split over two int boundaries, it will take two operations instead of one to access the variable plus you now have additional time spent working with the value in two pieces.

Your code above uses &obj1 to get the address of the struct variable. This will always work. But that was not your original question which was how to get the address of the struct variable when all you have is the address of a member. This you cannot reliably do. In fact, C++ adds hidden addresses inside the struct variable that are not declared in the struct.