What will be the output of the following C code?
#define sqr(x) x*x
main()
{
int a1;
a1=25/sqr(5);
printf("%d",a1);
}
Answer: A
The output of the code shown above is 25. This is because the arithmetic statement takes the form of: 25/5*5 (which is equal to 1). If the macro had been defined as #define sqr(x) (x*x), the output would have been 1.
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After preprocessing when the program is sent for compilation the macros are removed from the expanded source code.
Answer: A
Correct answer: TRUE
The preprocessor produces a translation unit in which every macro has been expanded and removed - no #define and no macro name survives into what the compiler parses. That is why a macro has no type, occupies no storage, and cannot be inspected in a debugger.
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What is #include directive?
Answer: A
Correct answer: Tells the preprocessor to grab the text of a file and place it directly into the current file
#include performs a textual insertion: the named file’s contents replace the directive, and the result is then processed as if it had been typed there. This is why a header can define macros and types that the including file goes on to use, and why include guards are needed to stop the same text arriving twice.
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What will be the output of the following C code?
#include <stdio.h>
#define a 2
main()
{
int r;
#define a 5
r=a*2;
printf("%d",r);
}
Answer: A
Correct answer: 10
The preprocessor works on the text in order and ignores the braces entirely, so by the time it reaches r = a*2; the macro a has been redefined to 5. The statement becomes r = 5*2;, giving 10. Expect a redefinition warning, since there is no #undef in between.
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What will be the output of the following C code?
#include <stdio.h>
#define max 100
main()
{
#ifdef max
printf("hello");
}
Answer: D
The code shown above results in an error. This is because the preprocessor #endif is missing in the above code. If the identifier is defined, then the statements following #ifdef are executed till #endif is encountered.
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A macro must always be defined in capital letters.
Answer: B
Correct answer: False
Capitals are a convention, not a requirement. They serve as a warning to the reader that the name will be substituted textually, with all the traps that carries. The preprocessor accepts any valid identifier, and the standard library itself defines lower-case macros such as getchar.
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What will be the output of the following C code?
#include <stdio.h>
#define max 100
void main()
{
#if(max)
printf("san");
#endif
printf("foundry");
}
Answer: D
The code shown above is an illustration of the preprocessor directive #if. The value of the defined identifier max is 100. On checking the condition (100510==), which is true, the statements after #if are executed till #endif is encountered. Hence the output is sanfoundry.
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What is a preprocessor directive
Answer: C
Correct answer: a message from programmer to the preprocessor
A directive such as #include, #define or #ifdef is an instruction to the preprocessor, the phase that transforms the source text before the compiler sees it. It is not part of the C language proper, which is why directives have their own syntax and end at the end of a line rather than at a semicolon.
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#include <stdio.h>
#define square(x) x*x
int main()
{
int x;
x = 36/square(6);
printf("%d", x);
return 0;
}
Answer: B
Correct answer: 36
The macro has no parentheses, so square(6) expands to the bare text 6*6 and the statement becomes
x = 36/6*6;
/ and * have equal precedence and associate left to right, so this is \((36/6) \times 6 = 6 \times 6 = 36\), not 1. Writing the macro as #define square(x) ((x)*(x)) is what prevents this.
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What will be the output of the program?
#include <stdio.h>
#define PRINT(i) printf("%d,",i)
int main()
{
int x=2, y=3, z=4;
PRINT(x);
PRINT(y);
PRINT(z);
return 0;
}
Answer: A
Correct answer: 2, 3, 4,
Each call substitutes its own argument into the macro body, so the three calls become printf("%d,", x), printf("%d,", y) and printf("%d,", z). The output is 2, 3, 4, - with a trailing comma, since the format string ends with one.
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