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- Question
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
- Correct Answer
- 3
ExplanationThe 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
C Preprocessor problems
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- 1. 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; }
Options- A. a = 10, b = 12
- B. a = 12, b = 10
- C. Error: Declaration not allowed in macro
- D. Error: Undefined symbol 't' Discuss
Correct Answer: a = 12, b = 10
Explanation:
The macro SWAP(a, b) int t; t=a, a=b, b=t; swaps the value of the given two variable.Step 1: int a=10, b=12; The variable a and b are declared as an integer type and initialized to 10, 12 respectively.
Step 2: SWAP(a, b);. Here the macro is substituted and it swaps the value to variable a and b.
Hence the output of the program is 12, 10.
- 2. What will be the output of the program?
#include<stdio.h> #define MAX(a, b) (a > b? a : b) int main() { int x; x = MAX(3+2, 2+7); printf("%d\n", x); return 0; }
Options- A. 8
- B. 9
- C. 6
- D. 5 Discuss
Correct Answer: 9
Explanation:
The macro MAX(a, b) (a > b ? a : b) returns the biggest value of the given two numbers.Step 1 : int x; The variable x is declared as an integer type.
Step 2 : x = MAX(3+2, 2+7); becomes,
=> x = (3+2 > 2+7 ? 3+2 : 2+7)
=> x = (5 > 9 ? 5 : 9)
=> x = 9
Step 3 : printf("%d\n", x); It prints the value of variable x.
Hence the output of the program is 9.
- 3. What will be the output of the program?
#include<stdio.h> #define str(x) #x #define Xstr(x) str(x) #define oper multiply int main() { char *opername = Xstr(oper); printf("%s\n", opername); return 0; }
Options- A. Error: in macro substitution
- B. Error: invalid reference 'x' in macro
- C. print 'multiply'
- D. No output Discuss
Correct Answer: print 'multiply'
Explanation:
The macro #define str(x) #x replaces the symbol 'str(x)' with 'x'.The macro #define Xstr(x) str(x) replaces the symbol 'Xstr(x)' with 'str(x)'.
The macro #define oper multiply replaces the symbol 'oper' with 'multiply'.
Step 1: char *opername = Xstr(oper); The varible *opername is declared as an pointer to a character type.
=> Xstr(oper); becomes,
=> Xstr(multiply);
=> str(multiply)
=> char *opername = multiply
Step 2: printf("%s\n", opername); It prints the value of variable opername.
Hence the output of the program is "multiply"
- 4. What will be the output of the program?
#include<stdio.h> #define FUN(arg) do\ {\ if(arg)\ printf("CuriousTab...", "\n");\ }while(--i) int main() { int i=2; FUN(i<3); return 0; }
Options- A. CuriousTab...
CuriousTab...
CuriousTab - B. CuriousTab... CuriousTab...
- C. Error: cannot use control instructions in macro
- D. No output Discuss
Correct Answer: CuriousTab... CuriousTab...
Explanation:
The macro FUN(arg) prints the statement "CuriousTab..." untill the while condition is satisfied.Step 1: int i=2; The variable i is declared as an integer type and initialized to 2.
Step 2: FUN(i<3); becomes,
do { if(2 < 3) printf("CuriousTab...", "\n"); }while(--2)After the 2 while loops the value of i becomes '0'(zero). Hence the while loop breaks.
Hence the output of the program is "CuriousTab... CuriousTab..."
- 5. 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="CURIOUSTAB"; JOIN(str1, str2); return 0; }
Options- A. str1=CuriousTab str2=CURIOUSTAB
- B. str1=India str2=CURIOUSTAB
- C. str1=India str2=CuriousTab
- D. Error: in macro substitution Discuss
Correct Answer: str1=India str2=CURIOUSTAB
- 6. 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.
- 7. 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.
- 8. 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
- 9. 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".
- 10. The preprocessor can trap simple errors like missing declarations, nested comments or mismatch of braces.
Options- A. True
- B. False Discuss
Correct Answer: False
Explanation:
False, the preprocessor cannot trap the errors, it only replaces the macro with the given expression. But the compiler will detect errors.
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- 1. What will be the output of the program?