Question
#include<stdio.h>
#define SQUARE(x) x*x
int main()
{
float s=10, u=30, t=2, a;
a = 2*(s-u*t)/SQUARE(t);
printf("Result = %f", a);
return 0;
}
What will be the output of the program?
#include<stdio.h>
#define SQUARE(x) x*x
int main()
{
float s=10, u=30, t=2, a;
a = 2*(s-u*t)/SQUARE(t);
printf("Result = %f", a);
return 0;
}
Answer: Option A
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The macro function SQUARE(x) x*x calculate the square of the given number 'x'. (Eg: 102)
Step 1: float s=10, u=30, t=2, a; Here the variable s, u, t, a are declared as an floating point
type and the variable s, u, t are initialized to 10, 30, 2.
Step 2: a = 2*(s-u*t)/SQUARE(t); becomes,
=> a = 2 * (10 - 30 * 2) / t * t; Here SQUARE(t) is replaced by macro to t*t .
=> a = 2 * (10 - 30 * 2) / 2 * 2;
=> a = 2 * (10 - 60) / 2 * 2;
=> a = 2 * (-50) / 2 * 2 ;
=> a = 2 * (-25) * 2 ;
=> a = (-50) * 2 ;
=> a = -100;
Step 3: printf("Result=%f", a); It prints the value of variable 'a'.
Hence the output of the program is -100
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