What will be the output of the program?

#include<stdio.h>
#define SQR(x)(x*x)

int main()
{
    int a, b=3;
    a = SQR(b+2);
    printf("%d\n", a);
    return 0;
}

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.

sizeof(3.14f) here '3.14f' specifies the float data type. Hence size of float is 4 bytes.

sizeof(3.14) here '3.14' specifies the double data type. Hence size of float is 8 bytes.

sizeof(3.14l) here '3.14l' specifies the long double data type. Hence size of float is 10 bytes.

Note: If you run the above program in Linux platform (GCC Compiler) it will give 4, 8, 12 as output. If you run in Windows platform (TurboC Compiler) it will give 4, 8, 10 as output. Because, C is a machine dependent language.

printf("%f\n", sqrt(36.0)); It prints the square root of 36 in the float format(i.e 6.000000).

Declaration Syntax: double sqrt(double x) calculates and return the positive square root of the given number.

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

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

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.