Which of the following is not a preprocessor directive?
Answer: D
Correct answer: #ifelse
There is no #ifelse directive. The conditional set is #if, #ifdef, #ifndef, #elif, #else and #endif - the “else if” form is spelt #elif. #error and #pragma are both real directives.
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Consider the following statements
#define hypotenuse(a, b) sqrt (a*a+b*b)
The macro call hypotenuse(a+2,b+3);
Answer: D
Correct answer: Find the square root of 3*a + 4*b + 5
The macro body has no parentheses round its parameters, so the arguments are pasted in as raw text:
sqrt(a+2*a+2 + b+3*b+3)
Multiplication binds tighter than addition, so this is \(\sqrt{a + 2a + 2 + b + 3b + 3} = \sqrt{3a + 4b + 5}\) - nothing like a hypotenuse. Writing #define hypotenuse(a, b) sqrt((a)*(a) + (b)*(b)) is what makes the macro behave as intended.
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What will be the output of the program?
#include <stdio.h>
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
int main()
{
char *str1="India";
char *str2="BIX";
JOIN(str1, str2);
return 0;
}
Answer: B
Correct answer: str1=India str2=BIX
Each parameter is used twice: once with #, which stringizes the argument’s text, and once plainly, which yields the pointer. So #s1 becomes "str1" while s1 is the variable holding "India". The output is str1=India str2=BIX - the idiom used by debugging macros that print both a variable’s name and its value.
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What is the output of this program?
#include <stdio.h>
#define i 5
int main()
{
#define i 10
printf("%d",i);
return 0;
}
Answer: B
Correct answer: 10
The second #define appears before the printf in the file, and the preprocessor works purely on text order - the braces of main mean nothing to it, so the redefinition is in force by the time the printf line is expanded. It prints 10, with a warning about the redefinition.
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_______________ is the preprocessor directive which is used to end the scope of #ifdef.
Answer: C
The #ifdef preprocessor directive is used to check if a particular identifier is currently defined or not. If the particular identifier is defined, the statements following this preprocessor directive are executed till another preprocessor directive #endif is encountered. #endif is used to end the scope of #ifdef.
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What will be the output of the following C code?
#include <stdio.h>
void f()
{
#define sf 100
printf("%d",sf);
}
int main()
{
#define sf 99;
f();
printf("%d",sf);
}
Answer: A
Correct answer: error
The macro body is 99; - the semicolon is part of the text. Substituting it gives printf("%d",99;);, a syntax error. (The redefinition of sf without an intervening #undef is a second fault.) A #define for a constant should never end in a semicolon.
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The preprocessor provides the ability for _______________
Answer: D
Correct answer: All of the mentioned
The preprocessor does all three: it handles file inclusion (#include), macro expansion (#define), and conditional compilation and line control (#if, #ifdef, #line). All of it happens on the text, before the compiler parses anything.
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What is the output of following program?
#include <stdio.h>
#define macro(n, a, i, m) m##a##i##n
#define MAIN macro(n, a, i, m)
int MAIN()
{
printf("GeeksQuiz");
return 0;
}
Answer: B
Correct answer: GeeksQuiz
## is the token-pasting operator. macro(n, a, i, m) substitutes the arguments and joins them in the order m a i n, producing the single token main. So int MAIN() becomes int main(), the program is well formed, and it prints GeeksQuiz.
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Will the program compile successfully?
#include <stdio.h>
int main()
{
#ifdef NOTE
int a;
a=10;
#else
int a;
a=20;
#endif
printf("%d\n", a);
return 0;
}
Answer: A
Correct answer: Yes
NOTE is not defined, so the #ifdef branch is deleted and only the #else branch reaches the compiler. Exactly one declaration of a therefore survives, and the program compiles and prints 20. Had both branches been kept, the duplicate declaration would have been an error - which is the point of the example.
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Point out the error in the program
#include <stdio.h>
int main()
{
int i;
#if A
printf("Enter any number:");
scanf("%d", &i);
#elif B
printf("The number is odd");
return 0;
}
Answer: A
Correct answer: Error: unexpected end of file because there is no matching #endif
Every #if, #ifdef or #ifndef must be closed by an #endif. Here the conditional started by #if A and continued by #elif B is never closed, so the preprocessor runs off the end of the file and reports it. Note that this is a preprocessing error, so it is reported before any of the C is looked at.
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