Q: What will be the output of the program? #include #define MESS junk int main() { printf("MESS\n"); return 0; }

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

Question 1

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
", a);
    return 0;
}

Accepted Answer

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
", a); It prints the value of variable 'a'. Hence the output of the program is 11

Question 2

What will be the output of the program? #include<stdio.h>
#define MESS junk

int main()
{
    printf("MESS
");
    return 0;
}

Accepted Answer

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

Question 3

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;
}

Accepted Answer

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

Question 4

What will be the output of the program? #include<stdio.h>
#define CUBE(x) (x*x*x)

int main()
{
    int a, b=3;
    a = CUBE(b++);
    printf("%d, %d
", a, b);
    return 0;
}

Accepted Answer

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
", a, b); It prints the value of variable a and b. Hence the output of the program is 27, 6.

Question 5

What will be the output of the program? #include<stdio.h>
#define PRINT(int) printf("int=%d, ", int);

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

Accepted Answer

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.

Question 6

What will be the output of the program? #include<stdio.h>
#define FUN(arg) do\
                 {\
                    if(arg)\
                        printf("CuriousTab...", "
");\
                  }while(--i)

int main()
{
    int i=2;…

Accepted Answer

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...", "
");
}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..."

Question 7

What will be the output of the program? #include<stdio.h>
#define str(x) #x
#define Xstr(x) str(x)
#define oper multiply

int main()
{
    char *opername = Xstr(oper);
    printf("%s
", opername);
    return 0;
}

Accepted Answer

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
", opername); It prints the value of variable opername. Hence the output of the program is "multiply"

Question 8

What will be the output of the program? #include<stdio.h>
#define MAX(a, b) (a > b? a : b)

int main()
{
    int x;
    x = MAX(3+2, 2+7);
    printf("%d
", x);
    return 0;
}

Accepted Answer

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
", x); It prints the value of variable x. Hence the output of the program is 9.

Question 9

What will be the output of the program? #include<stdio.h>
#define SWAP(a, b) int t; t=a, a=b, b=t;
int main()
{
    int a=10, b=12;
    SWAP(a, b);
    printf("a = %d, b = %d
", a, b);
    return 0;
}

Accepted Answer

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.

Question 10

What will be the output of the program? #include #define MIN(x, y) (x 0) printf("%d ", z); return 0; }

Accepted Answer

The macro MIN(x, y) (x 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 ", z);. It prints the value of variable z. Hence the output of the program is 3

Question 11

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;
}

Accepted Answer

The macro PRINT(i) print("%d,", i); 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("%d,",x). Hence it prints '2'. Step 3: PRINT(y); becomes printf("%d,",y). Hence it prints '3'. Step 4: PRINT(z); becomes printf("%d,",z). Hence it prints '4'. Hence the output of the program is 2, 3, 4.

Question 12

What will be the output of the program? #include<stdio.h>
#define FUN(i, j) i##j

int main()
{
    int va1=10;
    int va12=20;
    printf("%d
", FUN(va1, 2));
    return 0;
}

Accepted Answer

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
", FUN(First, Second) );

return 0;
}

Output:
-------
CuriousTab

The preprocessor will replace FUN(First, Second) as FirstSecond. Therefore, the printf("%s
", FUN(First, Second) ); statement will become as printf("%s
", FirstSecond ); Hence it prints CuriousTab as output. Like the same, the line printf("%d
", FUN(va1, 2)); given in the above question will become as printf("%d
", va12 );. Therefore, it prints 20 as output.

Question 13

What will be the output of the program? #include<stdio.h>
#define MAN(x, y) ((x)>(y))? (x):(y);

int main()
{
    int i=10, j=5, k=0;
    k = MAN(++i, j++);
    printf("%d, %d, %d
", i, j, k);
    return 0;
}

Accepted Answer

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
", 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

Question 14

What will be the output of the program? #include<stdio.h>
#define MAX(a, b, c) (a>b? a>c? a : c: b>c? b : c)

int main()
{
    int x;
    x = MAX(3+2, 2+7, 3+7);
    printf("%d
", x);
    return 0;
}

Accepted Answer

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
", x); It prints the value of 'x'. Hence the output of the program is "10".

Question 15

The preprocessor can trap simple errors like missing declarations, nested comments or mismatch of braces.

Accepted Answer

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

Question 16

A preprocessor directive is a message from programmer to the preprocessor.

Accepted Answer

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

Question 17

Preprocessor directive #ifdef .. #else ... #endif is used for conditional compilation.

Accepted Answer

True, these macros are used for conditional operation.

Question 18

Macro calls and function calls work exactly similarly.

Accepted Answer

False, A macro just replaces each occurrence with the code assigned to it. e.g. SQUARE(3) with ((3)*(3)) in the program. A function is compiled once and can be called from anywhere that has visibility to the funciton.

Question 19

If the file to be included doesn't exist, the preprocessor flashes an error message.

Accepted Answer

True, the included file does not exist it will generate the error.

C Preprocessor Questions