What will be the output of the program?

#include<stdio.h>
#define PRINT(i) printf("%d,",i)

int main()
{
    int x=2, y=3, z=4;
    PRINT(x);
    PRINT(y);
    PRINT(z);
    return 0;
}

The macro MIN(x, y) (x<y)? x : y; returns the smallest value from the given two numbers.

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

Step 2: z = MIN(x+y/2, y-1); becomes,

=> z = (x+y/2 < y-1)? x+y/2 : y - 1;

=> z = (3+4/2 < 4-1)? 3+4/2 : 4 - 1;

=> z = (3+2 < 4-1)? 3+2 : 4 - 1;

=> z = (5 < 3)? 5 : 3;

The macro return the number 3 and it is stored in the variable z.

Step 3: if(z > 0) becomes if(3 > 0) here the if condition is satisfied. It executes the if block statements.

Step 4: printf("%d\n", z);. It prints the value of variable z.

Hence the output of the program is 3

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 MAX(a, b) (a > b ? a : b) returns the biggest value of the given two numbers.

Step 1 : int x; The variable x is declared as an integer type.

Step 2 : x = MAX(3+2, 2+7); becomes,

=> x = (3+2 > 2+7 ? 3+2 : 2+7)

=> x = (5 > 9 ? 5 : 9)

=> x = 9

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

Hence the output of the program is 9.

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"

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 following program will make you understand about ## (macro concatenation) operator clearly.

#include<stdio.h>
#define FUN(i, j) i##j

int main()
{
    int First  	= 10;
    int Second  = 20;

    char FirstSecond[] = "CuriousTab";

    printf("%s\n", FUN(First, Second) );

    return 0;
}

Output:
-------
CuriousTab

The preprocessor will replace FUN(First, Second) as FirstSecond.

Therefore, the printf("%s\n", FUN(First, Second) ); statement will become as printf("%s\n", FirstSecond );

Hence it prints CuriousTab as output.

Like the same, the line printf("%d\n", FUN(va1, 2)); given in the above question will become as printf("%d\n", va12 );.

Therefore, it prints 20 as output.

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\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 macro MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c) returns the biggest of given three numbers.

Step 1: int x; The variable x is declared as an integer type.

Step 2: x = MAX(3+2, 2+7, 3+7); becomes,

=> x = (3+2 >2+7 ? 3+2 > 3+7 ? 3+2 : 3+7: 2+7 > 3+7 ? 2+7 : 3+7)

=> x = (5 >9 ? (5 > 10 ? 5 : 10): (9 > 10 ? 9 : 10) )

=> x = (5 >9 ? (10): (10) )

=> x = 10

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

Hence the output of the program is "10".

False, the preprocessor cannot trap the errors, it only replaces the macro with the given expression. But the compiler will detect errors.

True, the programmer tells the compiler to include the preprocessor when compiling.