Java C Preprocessor MCQ 2nd, Java C Preprocessor Question and Answers

Yes, this program will compile and run successfully and prints 20.

The macro #ifdef NOTE evaluates the given expression to 1. If satisfied it executes the #ifdef block statements. Here #ifdef condition fails because the Macro NOTE is nowhere declared.

Hence the #else block gets executed, the variable a is declared and assigned a value of 20.

printf(“%d “, a); It prints the value of variable a 20.

Because typedef goes to work after preprocessing.

Unless the header file has taken care to ensure that if already included it does not get included again.

Turbo C, GCC compilers would take care of these problems, generate no error.

False, Always the macro is substituted by the given text/expression.

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

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.

printf(“MESS “); 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 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
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.

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[] = “JavaTpoint”;

printf(“%s “, Fun(First, Second) );

return 0;
}

Output:
——-
JavaTpoint

The macro function SQR(x)(x*x) calculate the square of the given number x.
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

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 “, 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.

True, #undef can be used only on a macro that has been #define earlier
Example: #define PI 3.14
We can undefine PI macro by #undef PI

The macros #ifdef #if #elif are called conditional macros.
The macro #undef undefine the previosly declared macro symbol.
Hence all the given statements are macro preprocessor directives.

The conditional macro #if must have an #endif. In this program there is no #endif statement written.

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.