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- Question
What will be the output of the program?
#include<stdio.h> #define PRINT(int) printf("int=%d, ", int); int main() { int x=2, y=3, z=4; PRINT(x); PRINT(y); PRINT(z); return 0; }
Options- A. int=2, int=3, int=4
- B. int=2, int=2, int=2
- C. int=3, int=3, int=3
- D. int=4, int=4, int=4
- Correct Answer
- int=2, int=3, int=4
ExplanationThe macro PRINT(int) print("%d,", int); 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("int=%d,",x). Hence it prints 'int=2'.
Step 3: PRINT(y); becomes printf("int=%d,",y). Hence it prints 'int=3'.
Step 4: PRINT(z); becomes printf("int=%d,",z). Hence it prints 'int=4'.
Hence the output of the program is int=2, int=3, int=4.
C Preprocessor problems
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- 1. 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; }
Options- A. 9, 4
- B. 27, 4
- C. 27, 6
- D. Error Discuss
Correct Answer: 27, 6
Explanation:
The macro function CUBE(x) (x*x*x) calculates the cubic value of given number(Eg: 103.)Step 1: int a, b=3; The variable a and b are declared as an integer type and varaible b id initialized to 3.
Step 2: a = CUBE(b++); becomes
=> a = b++ * b++ * b++;
=> a = 3 * 3 * 3; Here we are using post-increement operator, so the 3 is not incremented in this statement.
=> a = 27; Here, 27 is store in the variable a. By the way, the value of variable b is incremented by 3. (ie: b=6)
Step 3: printf("%d, %d\n", a, b); It prints the value of variable a and b.
Hence the output of the program is 27, 6.
- 2. What will be the output of the program?
#include<stdio.h> #define SQUARE(x) x*x int main() { float s=10, u=30, t=2, a; a = 2*(s-u*t)/SQUARE(t); printf("Result = %f", a); return 0; }
Options- A. Result = -100.000000
- B. Result = -25.000000
- C. Result = 0.000000
- D. Result = 100.000000 Discuss
Correct Answer: Result = -100.000000
Explanation:
The macro function SQUARE(x) x*x calculate the square of the given number 'x'. (Eg: 102)Step 1: float s=10, u=30, t=2, a; Here the variable s, u, t, a are declared as an floating point type and the variable s, u, t are initialized to 10, 30, 2.
Step 2: a = 2*(s-u*t)/SQUARE(t); becomes,
=> a = 2 * (10 - 30 * 2) / t * t; Here SQUARE(t) is replaced by macro to t*t .
=> a = 2 * (10 - 30 * 2) / 2 * 2;
=> a = 2 * (10 - 60) / 2 * 2;
=> a = 2 * (-50) / 2 * 2 ;
=> a = 2 * (-25) * 2 ;
=> a = (-50) * 2 ;
=> a = -100;
Step 3: printf("Result=%f", a); It prints the value of variable 'a'.
Hence the output of the program is -100
- 3. What will be the output of the program?
#include<stdio.h> #define MESS junk int main() { printf("MESS\n"); return 0; }
Options- A. junk
- B. MESS
- C. Error
- D. Nothing will print Discuss
Correct Answer: MESS
Explanation:
printf("MESS\n"); It prints the text "MESS". There is no macro calling inside the printf statement occured.- 4. What will be the output of the program?
#include<stdio.h> #define SQR(x)(x*x) int main() { int a, b=3; a = SQR(b+2); printf("%d\n", a); return 0; }
Options- A. 25
- B. 11
- C. Error
- D. Garbage value Discuss
Correct Answer: 11
Explanation:
The macro function SQR(x)(x*x) calculate the square of the given number 'x'. (Eg: 102)Step 1: int a, b=3; Here the variable a, b are declared as an integer type and the variable b is initialized to 3.
Step 2: a = SQR(b+2); becomes,
=> a = b+2 * b+2; Here SQR(x) is replaced by macro to x*x .
=> a = 3+2 * 3+2;
=> a = 3 + 6 + 2;
=> a = 11;
Step 3: printf("%d\n", a); It prints the value of variable 'a'.
Hence the output of the program is 11
- 5. What will be the output of the program?
#include<stdio.h> int main() { float fval=7.29; printf("%d\n", (int)fval); return 0; }
Options- A. 0
- B. 0.0
- C. 7.0
- D. 7 Discuss
Correct Answer: 7
Explanation:
printf("%d\n", (int)fval); It prints '7'. because, we typecast the (int)fval in to integer. It converts the float value to the nearest integer value.- 6. 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
- 7. 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..."
- 8. 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"
- 9. 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.
- 10. 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.
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- 1. What will be the output of the program?