what happen to the Variable C?

Question

#include<stdio.h>
main()
{
    int a,b,c;
    printf("Enter two values\n");
    scanf("%d",&a);
    scanf("%d",&b);
    scanf("%d",&c);
    c=a+b;
    printf("the addition of %d and %d is %d",a,b,c);
}

i have given a=2;b=2;c=3;

the result is the addition of 2 and 2 is 4; but the value of c is declared by mine; but have given input value of c by knownly to check the response; but no where considered in equation and no error is coming why?

Answer 1

it is unclear what you are asking, but if you are wondering why c is equal to 4 it is because you declared it to be the value of a+b, which makes the scanf("%d",&c); line completely useless if I say 

int c=5;//initiates c with the value of 5

c=6;//changes that value to 6

printf("%d",c);

output:

6

so when you say

scanf("%d",&c);//is whatever the user inputs
    c=a+b;//is no longer user input, but instead a+b

you are overidding the initial assignment

Answer 2

why y have given the value to c?

because the value given to the variable to c is given error.........................but the last condition given to c is then changed to 4.....................                   .