Warning: implode(): Invalid arguments passed in /www/wwwroot/jobquiz.info/mdiscuss.php on line 336 What will be the output of the program? #include<stdio.h> #define SQUARE(x) xx int main() { float s=10, u=30, t=2, a; a = 2(s-ut)/SQUARE(t); printf("Result = %f", a); return 0; } ?->(Show Answer!)
1. What will be the output of the program? #include<stdio.h> #define SQUARE(x) xx int main() { float s=10, u=30, t=2, a; a = 2(s-ut)/SQUARE(t); printf("Result = %f", a); return 0; }
Ask Your Doubts Here
Comments
By: guest on 01 Jun 2017 06.00 pm
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