CODEWITHSUNDEEP
cpointersstruct
Amir Mohammad Nasrollahi
Amir Mohammad Nasrollahi

Reputation: 23

simple sum in structure with pointer in c

recently i started writing apps with c language. i have a problem with sum operator in below code. when i start program and i give for ex. 4 to int a; and 6 to int b; and then i want a+b witch give me 22. here is code, first i define my structure:

struct student {
   int a;
   int *b;
}*s1;

then i scan the values and the other steps:

int sum, x, y;
s1 = malloc(sizeof(struct student));

scanf("%d", &x);
scanf("%d", &y);

s1->a = x;
s1->b = y;
sum = (s1->a) + (s1->b);
printf("SUM: %d", sum);

Also i test *(s1->a) and still have problem. Thanks.

IMPORTANT: i don't want the other ways that i can solve this problem or suggesting the better ways of coding this algorithm, i just want to know why this piece of code not working properly and how can i fix it. Actually i want keep the 'b' as a pointer!

Upvotes: 0

Views: 2540

Answers (3)

Spikatrix
Spikatrix

Reputation: 20244

int *b in your structure is a pointer not an int. To fix your code,you can assign the address of y to the pointer like this:

s1->a = x;
s1->b = &y;
sum = (s1->a) + *(s1->b); //asterisk dereferences the pointer and gets the value stored in the memory address
printf("SUM: %d", sum);

Or just declare b as an int variable.

Upvotes: 0

Wintermute
Wintermute

Reputation: 44043

As @Mitch Wheat mentioned in the comments,

struct student {
   int a;
   int *b; // <-- this is a pointer, not an int
}*s1;

This has the effect that in the expression

(s1->a) + (s1->b);

pointer arithmetic is used, and the expression has the same value as &s1->b[s1->a] would have if s1->b were a valid pointer (that it isn't is a problem that is not triggered in the code).

Since s1->a == 4 and s1->b == memory address 6, the numerical value of (s1->a) + (s1->b) then is 6 + sizeof(int) * 4. If sizeof(int) == 4, this becomes 22, and that is the behavior you get.

The solution is to make b an int:

struct student {
   int a;
   int b;
}*s1;

Note that this is just a description of the behavior you get, not of behavior you are guaranteed to get. On a platform where pointers have a different size from ints, for example, the printf will not work as expected because the %d format specified does not use the whole pointer value.

Upvotes: 1

abligh
abligh

Reputation: 25129

Your problem is in the definition of the structure and/or how you are using it:

struct student {
   int a;
   int *b;
}*s1;

Here a is an int but b is a pointer to an int. If x had that struct type, you'd need to use x.a but *(x.b) to refer to the integer. So if s is a pointer to the struct, you need to use s->a but *(s->b).

However, you are using s1 as a pointer to the struct, but trying to use s->a and s->b seemingly both as integers. That won't work because of the way you have declared the struct.

More precisely:

s1->a = x;
s1->b = y;
sum = (s1->a) + (s1->b);
printf("SUM: %d", sum);

That sum line is doing pointer arithmetic (as s1->b is a pointer). This is unlikely to be what you required. Remove the * from the definition, i.e. use:

struct student {
   int a;
   int b;
} *s1;

You don't need to change the way you malloc() the struct. The fact that s1 is a pointer to the struct is a separate issue from whether the struct contains a pointer.

Upvotes: 0

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