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- Question
Size of short integer and long integer can be verified using the sizeof() operator.
Options- A. True
- B. False
- Correct Answer
- True
ExplanationTrue, we can find the size of short integer and long integer using the sizeof() operator.
Example:#include<stdio.h> int main() { short int i = 10; long int j = 10; printf("short int is %d bytes.,\nlong int is %d bytes.", sizeof(i),sizeof(j)); return 0; }Output:
short int is 2 bytes.
long int is 4 bytes.
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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