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- Question
When we mention the prototype of a function?
Options- A. Defining
- B. Declaring
- C. Prototyping
- D. Calling
- Correct Answer
- Declaring
ExplanationA function prototype in C or C++ is a declaration of a function that omits the function body but does specify the function's name, argument types and return type.While a function definition specifies what a function does, a function prototype can be thought of as specifying its interface.
More questions
- 1. Declare the following statement?
"A pointer to an array of three chars".
Options- A.
char *ptr[3](); - B.
char (*ptr)*[3]; - C.
char (*ptr[3])(); - D.
char (*ptr)[3];
Discuss
Correct Answer:char (*ptr)[3];
- 2. What will be the output of the program?
#include<stdio.h> int main() { unsigned int res; res = (64 >>(2+1-2)) & (~(1<<2)); printf("%d\n", res); return 0; }
Options- A. 32
- B. 64
- C. 0
- D. 128 Discuss
Correct Answer: 32
- 3. Union elements can be of different sizes.
Options- A. True
- B. False Discuss
Correct Answer: True
- 4. What will be the output of the program in DOS (Compiler - Turbo C)?
#include<stdio.h> double i; int main() { (int)(float)(char) i; printf("%d",sizeof(i)); return 0; }
Options- A. 4
- B. 8
- C. 16
- D. 22 Discuss
Correct Answer: 8
- 5. size of union is size of the longest element in the union
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
- 6. Bitwise & can be used to divide a number by powers of 2
Options- A. True
- B. False Discuss
Correct Answer: False
- 7. Assuming, integer is 2 byte, What will be the output of the program?
#include
; int main() { printf("%x\n", -1>>1); return 0; }
Options- A. ffff
- B. 0fff
- C. 0000
- D. fff0 Discuss
Correct Answer: ffff
Explanation:
Negative numbers are treated with 2's complement method.
1's complement: Inverting the bits ( all 1s to 0s and all 0s to 1s)
2's complement: Adding 1 to the result of 1's complement.
Binary of 1(2byte) : 0000 0000 0000 0001 Representing -1: 1s complement of 1(2byte) : 1111 1111 1111 1110 Adding 1 to 1's comp. result : 1111 1111 1111 1111 Right shift 1bit(-1>>1): 1111 1111 1111 1111 (carry out 1) Hexadecimal : f f f f (Filled with 1s in the left side in the above step)
Note:1. Fill with 1s in the left side for right shift for negative numbers.
2. Fill with 0s in the right side for left shift for negative numbers.
3. Fill with 0s in the left side for right shift for positive numbers.
4. Fill with 0s in the right side for left shift for positive numbers.- 8. What is the output of the program given below?
#include<stdio.h> int main() { enum status { pass, fail, atkt}; enum status stud1, stud2, stud3; stud1 = pass; stud2 = atkt; stud3 = fail; printf("%d, %d, %d\n", stud1, stud2, stud3); return 0; }
Options- A. 0, 1, 2
- B. 1, 2, 3
- C. 0, 2, 1
- D. 1, 3, 2 Discuss
Correct Answer: 0, 2, 1
Explanation:
enum takes the format like {0,1,2..) so pass=0, fail=1, atkt=2stud1 = pass (value is 0)
stud2 = atkt (value is 2)
stud3 = fail (value is 1)
Hence it prints 0, 2, 1
- 9. What will be the output of the program?
#include<stdio.h> int main() { char ch; ch = 'A'; printf("The letter is"); printf("%c", ch >= 'A' && ch <= 'Z'? ch + 'a' - 'A':ch); printf("Now the letter is"); printf("%c\n", ch >= 'A' && ch <= 'Z'? ch : ch + 'a' - 'A'); return 0; }
Options- A. The letter is a
Now the letter is A - B. The letter is A
Now the letter is a - C. Error
- D. None of above Discuss
Correct Answer: The letter is a
Now the letter is A
Explanation:
Step 1: char ch; ch = 'A'; here variable ch is declared as an character type an initialized to 'A'.Step 2: printf("The letter is"); It prints "The letter is".
Step 3: printf("%c", ch >= 'A' && ch <= 'Z' ? ch + 'a' - 'A':ch);
The ASCII value of 'A' is 65 and 'a' is 97.
Here
=> ('A' >= 'A' && 'A' <= 'Z') ? (A + 'a' - 'A'):('A')
=> (TRUE && TRUE) ? (65 + 97 - 65) : ('A')
=> (TRUE) ? (97): ('A')
In printf the format specifier is '%c'. Hence prints 97 as 'a'.
Step 4: printf("Now the letter is"); It prints "Now the letter is".
Step 5: printf("%c\n", ch >= 'A' && ch <= 'Z' ? ch : ch + 'a' - 'A');
Here => ('A' >= 'A' && 'A' <= 'Z') ? ('A') : (A + 'a' - 'A')
=> (TRUE && TRUE) ? ('A') :(65 + 97 - 65)
=> (TRUE) ? ('A') : (97)
It prints 'A'
Hence the output is
The letter is a
Now the letter is A- 10. What will be the output of the program?
#include<stdio.h> int main() { int k, num=30; k = (num>5? (num <=10? 100 : 200): 500); printf("%d\n", num); return 0; }
Options- A. 200
- B. 30
- C. 100
- D. 500 Discuss
Correct Answer: 30
Explanation:
Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to '30'.
Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement.
Step 3: printf("%d\n", num); It prints the value of variable num '30'
Step 3: Hence the output of the program is '30'
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- 1. Declare the following statement?