Which file is generated after pre-processing of a C program?
Answer: B
Correct answer: .i
The preprocessor produces a translation unit with all includes pulled in and all macros expanded, and by convention that file has the extension .i (.ii for C++). You can see it with gcc -E prog.c -o prog.i. .o is the object file produced later, by the assembler.
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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\n", a, b);
return 0;
}
Answer: C
Correct answer: 27, 6
CUBE(b++) expands to (b++*b++*b++), which modifies b three times in one expression with no sequence point between - formally undefined behaviour. The listed answer, 27 and 6, is what the common compilers give: the three operands all read 3 before any increment takes effect, and b ends three higher. Nothing should rely on it; the real lesson is never to pass an argument with a side effect to a macro that uses it more than once.
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What is the output of the following C program?
#include <stdio.h>
#define SQR(x) (x*x)
int main()
{
int a;
int b=4;
a=SQR(b+2);
printf("%d\n",a);
return 0;
}
Answer: A
Correct answer: 14
With b equal to 4, the unparenthesised macro expands to
a = b+2*b+2;
and multiplication binds tighter than addition: \(4 + 8 + 2 = 14\). The intended \(6^2 = 36\) requires the parameters to be parenthesised: #define SQR(x) ((x)*(x)).
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A macro must always be written in capital letters.
Answer: B
Correct answer: FALSE
Capitals are a strong convention, not a rule - they warn the reader that the name is a macro and will be substituted textually. The preprocessor accepts any valid identifier, and the standard library itself defines lower-case macros such as getchar and putc.
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It is necessary that a header files should have a .h extension?
Answer: B
Correct answer: No
The .h suffix is a convention. #include takes whatever name it is given and pastes in the contents of that file, so a header may be called anything - C++ standard headers have no extension at all. Following the convention is still worth doing, since it tells the reader what the file is for.
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What will be the output of the following C code?
#include <stdio.h>
#define max 20
main()
{
#ifndef max
#define min 10
#else
#define min 30
#endif
printf("%d",min);
}
Answer: C
Correct answer: 30
max is defined, so #ifndef max is false and the #else branch is taken, defining min as 30. The program prints 30.
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What is the output of this program?
#include <stdio.h>
#define int char
main()
{
int i=50;
printf ("sizeof (i) =%d", sizeof (i));
}
Answer: D
Correct answer: 1
#define int char replaces the token int everywhere it appears afterwards, so int i = 50; becomes char i = 50; and sizeof(i) is 1. The preprocessor has no notion of keywords being reserved - it substitutes tokens, and int is just a token to it.
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What is the output of this program?
#include <stdio.h>
#define CONDITION(x)
printf("letsfindcourse");
int main()
{
CONDITION(0);
return 0;
}
Answer: B
Correct answer: Compilation error
The macro CONDITION(x) has an empty body, since a #define ends at the end of its line. The printf below it is therefore an ordinary statement at file scope, where only declarations may appear, and the compiler rejects it. To make it part of the macro the first line would have to end in a backslash.
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If #include is used with file name in angular brackets.
Answer: A
Correct answer: The file is searched for in the standard compiler include paths
Angle brackets - #include <stdio.h> - tell the preprocessor to look in the standard system include directories. Quotes - #include "myheader.h" - add the directory of the current source file to the front of that search, which is why they are used for a project’s own headers.
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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\n", a, b);
return 0;
}
Answer: B
Correct answer: a = 12, b = 10
The macro expands to int t; t=a, a=b, b=t; inside main, where a declaration followed by an expression statement is perfectly legal. The comma operator sequences the three assignments left to right, so the values are exchanged and the output is a = 12, b = 10. The macro would break if used twice in the same block, since t would be declared twice.
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