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- Question
Will the following program print the message infinite number of times?
#include<stdio.h> #define INFINITELOOP while(1) int main() { INFINITELOOP printf("CuriousTab"); return 0; }
Options- A. Yes
- B. No
- Correct Answer
- Yes
ExplanationYes, the program prints "CuriousTab" and runs infinitely.The macro INFINITELOOP while(1) replaces the text 'INFINITELOOP' by 'while(1)'
In the main function, while(1) satisfies the while condition and it prints "CuriousTab". Then it comes to while(1) and the loop runs infinitely.
More questions
- 1. Bitwise can be used to perform addition and subtraction.
Options- A. Yes
- B. No Discuss
Correct Answer: No
- 2. Which of the following statements are correct about an if-else statements in a C-program?
1: Every if-else statement can be replaced by an equivalent statements using ?: operators 2: Nested if-else statements are allowed. 3: Multiple statements in an if block are allowed. 4: Multiple statements in an else block are allowed.
Options- A. 1 and 2
- B. 2 and 3
- C. 1, 2 and 4
- D. 2, 3, 4 Discuss
Correct Answer: 2, 3, 4
- 3. What will be the output of the program?
#include<stdio.h> int addmult(int ii, int jj) { int kk, ll; kk = ii + jj; ll = ii * jj; return (kk, ll); } int main() { int i=3, j=4, k, l; k = addmult(i, j); l = addmult(i, j); printf("%d, %d\n", k, l); return 0; }
Options- A. 12, 12
- B. 7, 7
- C. 7, 12
- D. 12, 7 Discuss
Correct Answer: 12, 12
Explanation:
Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively.The function addmult(i, j); accept 2 integer parameters.
Step 2: k = addmult(i, j); becomes k = addmult(3, 4)
In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'k'.
Step 3: l = addmult(i, j); becomes l = addmult(3, 4)
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'l'.
Step 4: printf("%d, %d\n", k, l); It prints the value of k and l
Hence the output is "12, 12".
- 4. What will be the output of the program?
#include<stdio.h> int main() { int i; char c; for(i=1; i<=5; i++) { scanf("%c", &c); /* given input is 'a' */ printf("%c", c); ungetc(c, stdin); } return 0; }
Options- A. aaaa
- B. aaaaa
- C. Garbage value.
- D. Error in ungetc statement. Discuss
Correct Answer: aaaaa
Explanation:
for(i=1; i<=5; i++) Here the for loop runs 5 times.Loop 1:
scanf("%c", &c); Here we give 'a' as input.
printf("%c", c); prints the character 'a' which is given in the previous "scanf()" statement.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.Loop 2:
Here the scanf("%c", &c); get the input from "stdin" because of "ungetc" function.
printf("%c", c); Now variable c = 'a'. So it prints the character 'a'.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.This above process will be repeated in Loop 3, Loop 4, Loop 5.
- 5. What will be the output of the program?
#include<stdio.h> int check(int); int main() { int i=45, c; c = check(i); printf("%d\n", c); return 0; } int check(int ch) { if(ch >= 45) return 100; else return 10; }
Options- A. 100
- B. 10
- C. 1
- D. 0 Discuss
Correct Answer: 100
Explanation:
Step 1: int check(int); This prototype tells the compiler that the function check() accepts one integer parameter and returns an integer value.Step 2: int l=45, c; The variable i and c are declared as an integer type and i is initialized to 45.
The function check(i) return 100 if the given value of variable i is >=(greater than or equal to) 45, else it will return 10.
Step 3: c = check(i); becomes c = check(45); The function check() return 100 and it get stored in the variable c.(c = 100)
Step 4: printf("%d\n", c); It prints the value of variable c.
Hence the output of the program is '100'.
- 6. Point out the error, if any in the for loop.
#include<stdio.h> int main() { int i=1; for(;;) { printf("%d\n", i++); if(i>10) break; } return 0; }
Options- A. There should be a condition in the for loop
- B. The two semicolons should be dropped
- C. The for loop should be replaced with while loop.
- D. No error Discuss
Correct Answer: No error
Explanation:
Step 1: for(;;) this statement will genereate infinite loop.
Step 2: printf("%d\n", i++); this statement will print the value of variable i and increement i by 1(one).
Step 3: if(i>10) here, if the variable i value is greater than 10, then the for loop breaks.
Hence the output of the program is
1
2
3
4
5
6
7
8
9
10
- 7. What will be the output of the program?
#include<stdio.h> int main() { float a=0.7; if(a < 0.7) printf("C\n"); else printf("C++\n"); return 0; }
Options- A. C
- B. C++
- C. Compiler error
- D. Non of above Discuss
Correct Answer: C
Explanation:
if(a < 0.7) here a is a float variable and 0.7 is a double constant. The float variable a is less than double constant 0.7. Hence the if condition is satisfied and it prints 'C'
Example:#include<stdio.h> int main() { float a=0.7; printf("%.10f %.10f\n",0.7, a); return 0; }Output:
0.7000000000 0.6999999881- 8. Assuming, integer is 2 byte, What will be the output of the program?
#include<stdio.h> int main() { printf("%x\n", -2<<2); return 0; }
Options- A. ffff
- B. 0
- C. fff8
- D. Error Discuss
Correct Answer: fff8
Explanation:
The integer value 2 is represented as 00000000 00000010 in binary system.
Negative numbers are represented in 2's complement method.
1's complement of 00000000 00000010 is 11111111 11111101 (Change all 0s to 1 and 1s to 0).
2's complement of 00000000 00000010 is 11111111 11111110 (Add 1 to 1's complement to obtain the 2's complement value).
Therefore, in binary we represent -2 as: 11111111 11111110.
After left shifting it by 2 bits we obtain: 11111111 11111000, and it is equal to "fff8" in hexadecimal system.- 9. What will be the output of the program?
#include<stdio.h> int main() { int i; i = printf("How r u\n"); i = printf("%d\n", i); printf("%d\n", i); return 0; }
Options- A. How r u
7
2 - B. How r u
8
2 - C. How r u
1
1 - D. Error: cannot assign printf to variable Discuss
Correct Answer: How r u
8
2
Explanation:
In the program, printf() returns the number of charecters printed on the console i = printf("How r u\n"); This line prints "How r u" with a new line character and returns the length of string printed then assign it to variable i.
So i = 8 (length of '\n' is 1).i = printf("%d\n", i); In the previous step the value of i is 8. So it prints "8" with a new line character and returns the length of string printed then assign it to variable i. So i = 2 (length of '\n' is 1).
printf("%d\n", i); In the previous step the value of i is 2. So it prints "2".
- 10. Will the program compile successfully?
#include<stdio.h> int main() { #ifdef NOTE int a; a=10; #else int a; a=20; #endif printf("%d\n", a); return 0; }
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
Explanation:
Yes, this program will compile and run successfully and prints 20.The macro #ifdef NOTE evaluates the given expression to 1. If satisfied it executes the #ifdef block statements. Here #ifdef condition fails because the Macro NOTE is nowhere declared.
Hence the #else block gets executed, the variable a is declared and assigned a value of 20.
printf("%d\n", a); It prints the value of variable a 20.
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