The macro SWAP(a, b) int t; t=a, a=b, b=t; swaps the value of the given two variable.

Step 1: int a=10, b=12; The variable a and b are declared as an integer type and initialized 

to 10, 12 respectively.

Step 2: SWAP(a, b);. Here the macro is substituted and it swaps the value to variable a and b.

Hence the output of the program is 12, 10.

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 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`setminus`n", a, b); It prints the value of variable a and b.

Hence the output of the program is 27, 6.

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`setminus`n", a); It prints the value of variable 'a'.

Hence the output of the program is 11

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 MAN(x, y) ((x)>(y)) ? (x):(y); returns the biggest number of given two numbers.

Step 1: int i=10, j=5, k=0; The variable i, j, k are declared as an integer type and initialized to

 value 10, 5, 0 respectively.

Step 2: k = MAN(++i, j++); becomes,

=> k = ((++i)>(j++)) ? (++i):(j++);

=> k = ((11)>(5)) ? (12):(6);

=> k = 12

Step 3: printf("%d, %d, %d`setminus`n", i, j, k); It prints the variable i, j, k.

In the above macro step 2 the variable i value is increemented by 2 and variable j value is 

increemented by 1.

Hence the output of the program is 12, 6, 12

The preprocessor replaces the line #include <stdio.h> with the system header file of 

that name. More precisely, the entire text of the file 'stdio.h' replaces the #include directive.

The code won't compile since declaration of t cannot occur within parenthesis.

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, %dn", a, b); It prints the value of variable a and b.

Hence the output of the program is 27, 6.

As we know #define is a token pasting preprocessor it only paste the value of micro constant in the program, before the actual compilation start.
So pasting 5+2 in place of X, we have
i = 5+2*5+2*5+2
 = 5+10+10+2
= 27