C Preprocessor Questions and Answers

Q: #include <stdio.h> #define a 10 int main() { printf("%d ",a); #define a 50 printf("%d ",a); return 0; }

Correct answer: 10 50The preprocessor works line by line, top to bottom, so the first printf is expanded while a still means 10 and the second after it has been redefined to 50. The output is 10 50. Redefining a macro to a different value without an intervening #undef is not strictly conforming, so a compiler will warn.

Q: What is the output of this program? #include <stdio.h> int main() { printf("%d",30); return 0; }

Correct answer: 30There is nothing wrong with the program: <stdio.h> is included, main returns int, and printf("%d", 30) matches its format specifier to an int argument. It compiles cleanly and prints 30.

Q: In a macro call the control is passed to the macro.

Correct answer: FalseThere is no transfer of control. A macro is expanded in place, before compilation, so its body ends up written out at each call site - no jump, no stack frame, no return. That is the essential difference from a function: the speed of inline code, at the cost of size and of type checking.

Q: Will the following program print the message infinite number of times? #include <stdio.h> #define INFINITELOOP while(1) int main() { INFINITELOOP printf("IndiaBIX"); return 0; }

Correct answer: YesThe macro expands to while(1), and the printf on the next line becomes the loop’s body - a while without braces takes the single statement that follows it, and a newline means nothing to the compiler. So the message is printed forever, and return 0; is never reached.

Q: 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\n", FUN(va1, 2)); return 0; }

Correct answer: 20## pastes its operands into a single token, so FUN(va1, 2) becomes va12 - the name of the second variable. The printf therefore prints 20.

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

Correct answer: True#ifdef / #else / #endif is exactly the mechanism for conditional compilation: only one branch survives preprocessing, and the other never reaches the compiler at all. It is the usual way of adapting a source file to different platforms or of compiling out debugging code.

Q: A preprocessor directive is a message from compiler to a linker.

Correct answer: FalseA directive is a message from the programmer to the preprocessor. The compiler communicates with the linker through the object file it produces, not through anything written with a #.

Q: In which stage the following code #include <stdio.h> gets replaced by the contents of the file stdio.h

Correct answer: During preprocessing#include is handled in the preprocessing phase, before compilation: the header’s text is pasted in, producing a single translation unit for the compiler. Nothing about it happens at link time or run time.

Q: What will be the output of the following C code? #include <stdio.h> #define hello 10 void main() { printf("%d",hello); #undef hello printf("%d",hello); }

Correct answer: errorThe first printf is fine, but #undef hello removes the definition, so in the second printf the identifier hello is no longer replaced by 10. It is then an undeclared variable, and the compiler reports an error.

31.

#include <stdio.h>
#define a 10
int main()
{
  printf("%d ",a);
 
  #define a 50
 
  printf("%d ",a);
  return 0;
}

A.Compiler Error
B.10 50
C.50 50
D.10 10

Answer: B

Correct answer: 10 50

The preprocessor works line by line, top to bottom, so the first printf is expanded while a still means 10 and the second after it has been redefined to 50. The output is 10 50. Redefining a macro to a different value without an intervening #undef is not strictly conforming, so a compiler will warn.

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32.

What is the output of this program?

#include <stdio.h>
int main()
{
  printf("%d",30);
  return 0;
}

A.Runtime Error
B.Garbage value
C.Compilation error
D.30

Answer: C

Correct answer: 30

There is nothing wrong with the program: <stdio.h> is included, main returns int, and printf("%d", 30) matches its format specifier to an int argument. It compiles cleanly and prints 30.

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33.

In a macro call the control is passed to the macro.

A.True
B.False

Answer: B

Correct answer: False

There is no transfer of control. A macro is expanded in place, before compilation, so its body ends up written out at each call site - no jump, no stack frame, no return. That is the essential difference from a function: the speed of inline code, at the cost of size and of type checking.

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34.

Will the following program print the message infinite number of times?

#include <stdio.h>
#define INFINITELOOP while(1)

int main()
{
    INFINITELOOP
    printf("IndiaBIX");
    return 0;
}

A.Yes
B.No

Answer: A

Correct answer: Yes

The macro expands to while(1), and the printf on the next line becomes the loop’s body - a while without braces takes the single statement that follows it, and a newline means nothing to the compiler. So the message is printed forever, and return 0; is never reached.

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35.

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\n", FUN(va1, 2));
    return 0;
}

A.10
B.20
C.1020
D.12

Answer: B

Correct answer: 20

## pastes its operands into a single token, so FUN(va1, 2) becomes va12 - the name of the second variable. The printf therefore prints 20.

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36.

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

A.True
B.False

Answer: A

Correct answer: True

#ifdef / #else / #endif is exactly the mechanism for conditional compilation: only one branch survives preprocessing, and the other never reaches the compiler at all. It is the usual way of adapting a source file to different platforms or of compiling out debugging code.

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37.

A preprocessor directive is a message from compiler to a linker.

A.True
B.False

Answer: B

Correct answer: False

A directive is a message from the programmer to the preprocessor. The compiler communicates with the linker through the object file it produces, not through anything written with a #.

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38.

What will be the output of the following C code?

#include <stdio.h>
#define hello 
main()
{
    #ifdef hello
    #define hi 4
    #else
    #define hi 5
    #endif
    printf("%d",hi);
 
}

A.4
B.5
C.45
D.error

Answer: A

The code shown above illustrates #define, #ifdef, #else, #endif. Since the identifier hello is defined (condition of #if is true), another identifier names hi is defined and it’s value is 4. If hello was not defined, then the value of hi would be 5.

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39.

In which stage the following code

#include <stdio.h>

gets replaced by the contents of the file stdio.h

A.During editing
B.During linking
C.During execution
D.During preprocessing

Answer: D

Correct answer: During preprocessing

#include is handled in the preprocessing phase, before compilation: the header’s text is pasted in, producing a single translation unit for the compiler. Nothing about it happens at link time or run time.

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40.

What will be the output of the following C code?

#include <stdio.h>
#define hello 10
void main()
{
    printf("%d",hello);
    #undef hello
    printf("%d",hello);
}

A.10
B.hello
C.error
D.1010

Answer: C

Correct answer: error

The first printf is fine, but #undef hello removes the definition, so in the second printf the identifier hello is no longer replaced by 10. It is then an undeclared variable, and the compiler reports an error.

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