What will be the output of the program?

#include<stdio.h>
#include<math.h>
int main()
{
    printf("%d, %d, %d\n", sizeof(3.14f), sizeof(3.14), sizeof(3.14l));
    return 0;
}

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.

if(a < 0.7) here a is a float variable and 0.7 is a double constant. The float variable a is less than double constant 0.7. Hence the if condition is satisfied and it prints 'C'
Example:

#include<stdio.h>
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7, a);
    return 0;
}

Output:
0.7000000000 0.6999999881

if(a < 0.7f) here a is a float variable and 0.7f is a float constant. The float variable a is not less than 0.7f float constant. But both are equal. Hence the if condition is failed and it goes to else it prints 'C++'
Example:

#include<stdio.h>
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7f, a);
    return 0;
}

Output:
0.6999999881 0.6999999881

sizeof(x) returns the size of x in bytes.
float *p is a pointer to a float.

In 16 bit compiler, the pointer size is always 2 bytes.
In 32 bit compiler, the pointer size is always 4 bytes.

printf("%e,", d); Here '%e' specifies the "Scientific Notation" format. So, it prints the 2.25 as 2.250000e⁺⁰⁰⁰.

printf("%f,", d); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 2.25 as 2.250000.

printf("%g,", d); Here '%g' "Use the shorter of %e or %f". So, it prints the 2.25 as 2.25.

printf("%lf,", d); Here '%lf' specifies the "Long Double" format. So, it prints the 2.25 as 2.250000.

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.

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.

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.

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