Identify the macro(s) defined in stdarg.h.
Answer: D
Correct answer: all the above
<stdarg.h> defines all three: va_start to begin walking the variable argument list, va_arg to fetch the next argument (given its type), and va_end to finish. They are what a function such as printf uses to reach arguments the compiler cannot type-check.
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What will be the output of the following C code?
#include <stdio.h>
#define sf 10
main()
{
if(sf==100)
printf("good");
else
{
printf("bad");
sf=100;
}
printf("%d",sf);
}
Answer: D
The above code will result in an error because a macro cannot be defined using a simple assignment operator. In the code shown above, the value of sf is changed to 100 using the assignment operator. This causes an error.
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Undefined function calls in a C program are detected
Answer: C
Correct answer: by the linker
The compiler is satisfied by a declaration: it only needs to know the function’s signature to generate a call. Whether a definition actually exists is settled when the object files and libraries are combined, so a missing function surfaces as a linker error - the familiar “undefined reference”.
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There exists a way to prevent the same file from getting #included twice in the same program.
Answer: A
Correct answer: True
That is what an include guard does. The header wraps its contents in #ifndef HDR_H / #define HDR_H / #endif, so on any later inclusion the macro is already defined and the body is skipped. #pragma once achieves the same thing in one line, though it is a compiler extension rather than standard C.
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Macros have a local scope.
Answer: B
Correct answer: False
A macro has no scope in the C sense at all. Its definition runs from the #define to the end of the file, or to a matching #undef - braces and functions mean nothing to the preprocessor, which works on text. A #define written inside a function is still in force after that function ends.
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What will the SWAP macro in the following program be expanded to on preprocessing? will the code compile?
#include <stdio.h>
#define SWAP(a, b, c)(c t; t=a, a=b, b=t)
int main()
{
int x=10, y=20;
SWAP(x, y, int);
printf("%d %d\n", x, y);
return 0;
}
Answer: C
Correct answer: Not compile
The macro expands to
(int t; t=x, x=y, y=t);
which puts a declaration inside a parenthesised expression - something C does not allow. The trick of passing a type as a macro argument works only if the expansion places the declaration where a declaration may appear, typically inside a do { … } while(0) block.
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#include <stdio.h>
#define ISEQUAL(X, Y) X == Y
int main()
{
#if ISEQUAL(X, 0)
printf("Geeks");
#else
printf("Quiz");
#endif
return 0;
}
Output of the above program?
Answer: A
Correct answer: Geeks
The macro expands to X == 0, and inside a #if the undefined identifier X becomes 0. The condition is therefore 0 == 0, which is true, so printf("Geeks") is the line that survives preprocessing.
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#include statement must be written __________
Answer: B
Correct answer: Before any scanf/printf
#include <stdio.h> supplies the declarations of printf and scanf, and a function must be declared before it is called. So the include has to come first - which in practice means at the top of the file, above main.
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Preprocessor directive #undef can be used only on a macro that has been #define earlier
Answer: A
Correct answer: True
#undef does one thing: it removes an existing macro definition. Applied to a name that was never defined it has nothing to remove and is simply ignored, so the directive is meaningful only for a macro defined earlier. Its usual purpose is to redefine a macro safely - #undef then #define - or to stop a macro shadowing a real function name.
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Point out the error in the program
#include <stdio.h>
#define SI(p, n, r) float si; si=p*n*r/100;
int main()
{
float p=2500, r=3.5;
int n=3;
SI(p, n, r);
SI(1500, 2, 2.5);
return 0;
}
Answer: C
Correct answer: Error: Multiple declaration of si
The macro body declares si, so using it twice in the same block expands to two float si; declarations of the same name in the same scope - which the compiler rejects. This is the standard argument for wrapping a multi-statement macro in do { … } while(0), which gives each expansion its own block and its own scope.
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