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;
}

printf("MESS\n"); It prints the text "MESS". There is no macro calling inside the printf statement occured.

The macro function SQR(x)(x*x) calculate the square of the given number 'x'. (Eg: 10²)

Step 1: int a, b=3; Here the variable a, b are declared as an integer type and the variable b is initialized to 3.

Step 2: a = SQR(b+2); becomes,

=> a = b+2 * b+2; Here SQR(x) is replaced by macro to x*x .

=> a = 3+2 * 3+2;

=> a = 3 + 6 + 2;

=> a = 11;

Step 3: printf("%d\n", a); It prints the value of variable 'a'.

Hence the output of the program is 11

printf("%d\n", (int)fval); It prints '7'. because, we typecast the (int)fval in to integer. It converts the float value to the nearest integer value.

printf("%e, ", f); Here '%e' specifies the "Scientific Notation" format. So, it prints the 43.20 as 4.320000e⁺⁰¹.

printf("%f, ", f); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 43.20 as 43.200001.

printf("%g, ", f); Here '%g' "Use the shorter of %e or %f". So, it prints the 43.20 as 43.2.

ceil(x) round up the given value. It finds the smallest integer not < x.
floor(x) round down the given value. It finds the smallest integer not > x.

printf("%f, %f\n", ceil(n), floor(n)); In this line ceil(1.54) round up the 1.54 to 2 and floor(1.54) round down the 1.54 to 1.

In the printf("%f, %f\n", ceil(n), floor(n)); statement, the format specifier "%f %f" tells output to be float value. Hence it prints 2.000000 and 1.000000.

The macro function CUBE(x) (x*x*x) calculates the cubic value of given number(Eg: 10³.)

Step 1: int a, b=3; The variable a and b are declared as an integer type and varaible b id initialized to 3.

Step 2: a = CUBE(b++); becomes

=> a = b++ * b++ * b++;

=> a = 3 * 3 * 3; Here we are using post-increement operator, so the 3 is not incremented in this statement.

=> a = 27; Here, 27 is store in the variable a. By the way, the value of variable b is incremented by 3. (ie: b=6)

Step 3: printf("%d, %d\n", a, b); It prints the value of variable a and b.

Hence the output of the program is 27, 6.

The macro PRINT(int) print("%d,", int); prints the given variable value in an integer format.

Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.

Step 2: PRINT(x); becomes printf("int=%d,",x). Hence it prints 'int=2'.

Step 3: PRINT(y); becomes printf("int=%d,",y). Hence it prints 'int=3'.

Step 4: PRINT(z); becomes printf("int=%d,",z). Hence it prints 'int=4'.

Hence the output of the program is int=2, int=3, int=4.

The macro FUN(arg) prints the statement "CuriousTab..." untill the while condition is satisfied.

Step 1: int i=2; The variable i is declared as an integer type and initialized to 2.

Step 2: FUN(i<3); becomes,

do
{
    if(2 < 3)
    printf("CuriousTab...", "\n");
}while(--2)

After the 2 while loops the value of i becomes '0'(zero). Hence the while loop breaks.

Hence the output of the program is "CuriousTab... CuriousTab..."

The macro #define str(x) #x replaces the symbol 'str(x)' with 'x'.

The macro #define Xstr(x) str(x) replaces the symbol 'Xstr(x)' with 'str(x)'.

The macro #define oper multiply replaces the symbol 'oper' with 'multiply'.

Step 1: char *opername = Xstr(oper); The varible *opername is declared as an pointer to a character type.

=> Xstr(oper); becomes,

=> Xstr(multiply);

=> str(multiply)

=> char *opername = multiply

Step 2: printf("%s\n", opername); It prints the value of variable opername.

Hence the output of the program is "multiply"