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- Question
The preprocessor can trap simple errors like missing declarations, nested comments or mismatch of braces.
Options- A. True
- B. False
- Correct Answer
- False
ExplanationFalse, the preprocessor cannot trap the errors, it only replaces the macro with the given expression. But the compiler will detect errors.
C Preprocessor problems
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- 1. What will be the output of the program?
#include<stdio.h> #define MAX(a, b, c) (a>b? a>c? a : c: b>c? b : c) int main() { int x; x = MAX(3+2, 2+7, 3+7); printf("%d\n", x); return 0; }
Options- A. 5
- B. 9
- C. 10
- D. 3+7 Discuss
Correct Answer: 10
Explanation:
The macro MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c) returns the biggest of given three numbers.Step 1: int x; The variable x is declared as an integer type.
Step 2: x = MAX(3+2, 2+7, 3+7); becomes,
=> x = (3+2 >2+7 ? 3+2 > 3+7 ? 3+2 : 3+7: 2+7 > 3+7 ? 2+7 : 3+7)
=> x = (5 >9 ? (5 > 10 ? 5 : 10): (9 > 10 ? 9 : 10) )
=> x = (5 >9 ? (10): (10) )
=> x = 10
Step 3: printf("%d\n", x); It prints the value of 'x'.
Hence the output of the program is "10".
- 2. What will be the output of the program?
#include<stdio.h> #define MAN(x, y) ((x)>(y))? (x):(y); int main() { int i=10, j=5, k=0; k = MAN(++i, j++); printf("%d, %d, %d\n", i, j, k); return 0; }
Options- A. 12, 6, 12
- B. 11, 5, 11
- C. 11, 5, Garbage
- D. 12, 6, Garbage Discuss
Correct Answer: 12, 6, 12
Explanation:
The macro MAN(x, y) ((x)>(y)) ? (x):(y); returns the biggest number of given two numbers.Step 1: int i=10, j=5, k=0; The variable i, j, k are declared as an integer type and initialized to value 10, 5, 0 respectively.
Step 2: k = MAN(++i, j++); becomes,
=> k = ((++i)>(j++)) ? (++i):(j++);
=> k = ((11)>(5)) ? (12):(6);
=> k = 12
Step 3: printf("%d, %d, %d\n", i, j, k); It prints the variable i, j, k.
In the above macro step 2 the variable i value is increemented by 2 and variable j value is increemented by 1.
Hence the output of the program is 12, 6, 12
- 3. 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; }
Options- A. 10
- B. 20
- C. 1020
- D. 12 Discuss
Correct Answer: 20
Explanation:
The following program will make you understand about ## (macro concatenation) operator clearly.#include<stdio.h> #define FUN(i, j) i##j int main() { int First = 10; int Second = 20; char FirstSecond[] = "CuriousTab"; printf("%s\n", FUN(First, Second) ); return 0; } Output: ------- CuriousTabThe preprocessor will replace FUN(First, Second) as FirstSecond.
Therefore, the printf("%s\n", FUN(First, Second) ); statement will become as printf("%s\n", FirstSecond );
Hence it prints CuriousTab as output.
Like the same, the line printf("%d\n", FUN(va1, 2)); given in the above question will become as printf("%d\n", va12 );.
Therefore, it prints 20 as output.
- 4. 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; }
Options- A. 2, 3, 4,
- B. 2, 2, 2,
- C. 3, 3, 3,
- D. 4, 4, 4, Discuss
Correct Answer: 2, 3, 4,
Explanation:
The macro PRINT(i) print("%d,", i); prints the given variable value in an integer format.Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.
Step 2: PRINT(x); becomes printf("%d,",x). Hence it prints '2'.
Step 3: PRINT(y); becomes printf("%d,",y). Hence it prints '3'.
Step 4: PRINT(z); becomes printf("%d,",z). Hence it prints '4'.
Hence the output of the program is 2, 3, 4.
- 5. What will be the output of the program?
#include<stdio.h> #define MIN(x, y) (x<y)? x : y; int main() { int x=3, y=4, z; z = MIN(x+y/2, y-1); if(z > 0) printf("%d\n", z); return 0; }
Options- A. 3
- B. 4
- C. 0
- D. No output Discuss
Correct Answer: 3
Explanation:
The macro MIN(x, y) (x<y)? x : y; returns the smallest value from the given two numbers.Step 1: int x=3, y=4, z; The variable x, y, z are declared as an integer type and the variable x, y are initialized to value 3, 4 respectively.
Step 2: z = MIN(x+y/2, y-1); becomes,
=> z = (x+y/2 < y-1)? x+y/2 : y - 1;
=> z = (3+4/2 < 4-1)? 3+4/2 : 4 - 1;
=> z = (3+2 < 4-1)? 3+2 : 4 - 1;
=> z = (5 < 3)? 5 : 3;
The macro return the number 3 and it is stored in the variable z.
Step 3: if(z > 0) becomes if(3 > 0) here the if condition is satisfied. It executes the if block statements.
Step 4: printf("%d\n", z);. It prints the value of variable z.
Hence the output of the program is 3
- 6. A preprocessor directive is a message from programmer to the preprocessor.
Options- A. True
- B. False Discuss
Correct Answer: True
Explanation:
True, the programmer tells the compiler to include the preprocessor when compiling.- 7. Preprocessor directive #ifdef .. #else ... #endif is used for conditional compilation.
Options- A. True
- B. False Discuss
Correct Answer: True
Explanation:
True, these macros are used for conditional operation.- 8. Macro calls and function calls work exactly similarly.
Options- A. True
- B. False Discuss
Correct Answer: False
Explanation:
False, A macro just replaces each occurrence with the code assigned to it. e.g. SQUARE(3) with ((3)*(3)) in the program.A function is compiled once and can be called from anywhere that has visibility to the funciton.
- 9. If the file to be included doesn't exist, the preprocessor flashes an error message.
Options- A. True
- B. False Discuss
Correct Answer: True
Explanation:
True, the included file does not exist it will generate the error.- 10. Preprocessor directive #undef can be used only on a macro that has been #define earlier
Options- A. True
- B. False Discuss
Correct Answer: True
Explanation:
True, #undef can be used only on a macro that has been #define earlierExample: #define PI 3.14
We can undefine PI macro by #undef PI
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- 1. What will be the output of the program?