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- Question
Point out the error in the program?
#include<stdio.h> #include<string.h> void modify(struct emp*); struct emp { char name[20]; int age; }; int main() { struct emp e = {"Sanjay", 35}; modify(&e); printf("%s %d", e.name, e.age); return 0; } void modify(struct emp *p) { p ->age=p->age+2; }
Options- A. Error: in structure
- B. Error: in prototype declaration unknown struct emp
- C. No error
- D. None of above
- Correct Answer
- Error: in prototype declaration unknown struct emp
ExplanationThe struct emp is mentioned in the prototype of the function modify() before declaring the structure.To solve this problem declare struct emp before the modify() prototype. - 1. Bitwise & can be used to check if a bit in number is set or not.
Options- A. True
- B. False Discuss
- 2. A pointer union CANNOT be created
Options- A. Yes
- B. No Discuss
- 3. What will be the output of the program?
#include<stdio.h> #include<stdlib.h> int main() { union test { int i; float f; char c; }; union test *t; t = (union test *)malloc(sizeof(union test)); t->f = 10.10f; printf("%f", t->f); return 0; }
Options- A. 10
- B. Garbage value
- C. 10.100000
- D. Error Discuss
- 4. What will be the output of the program?
#include<stdio.h> int main() { int x=4, y, z; y = --x; z = x--; printf("%d, %d, %d\n", x, y, z); return 0; }
Options- A. 4, 3, 3
- B. 4, 3, 2
- C. 3, 3, 2
- D. 2, 3, 3 Discuss
- 5. Functions cannot return a floating point number
Options- A. Yes
- B. No Discuss
- 6. What will be the output of the program?
#include<stdio.h> int main() { int a=100, b=200, c; c = (a == 100 || b > 200); printf("c=%d\n", c); return 0; }
Options- A. c=100
- B. c=200
- C. c=1
- D. c=300 Discuss
- 7. What will be the output of the program?
#include<stdio.h> int main() { int x=55; printf("%d, %d, %d\n", x<=55, x=40, x>=10); return 0; }
Options- A. 1, 40, 1
- B. 1, 55, 1
- C. 1, 55, 0
- D. 1, 1, 1 Discuss
- 8. What will be the output of the program?
#include<stdio.h> int main() { void fun(char*); char a[100]; a[0] = 'A'; a[1] = 'B'; a[2] = 'C'; a[3] = 'D'; fun(&a[0]); return 0; } void fun(char *a) { a++; printf("%c", *a); a++; printf("%c", *a); }
Options- A. AB
- B. BC
- C. CD
- D. No output Discuss
- 9. Which of the following statements correct about k used in the below statement?
char ****k;
Options- A. k is a pointer to a pointer to a pointer to a char
- B. k is a pointer to a pointer to a pointer to a pointer to a char
- C. k is a pointer to a char pointer
- D. k is a pointer to a pointer to a char Discuss
- 10. What will be the output of the program?
#include<stdio.h> void fun(void *p); int i; int main() { void *vptr; vptr = &i; fun(vptr); return 0; } void fun(void *p) { int **q; q = (int**)&p; printf("%d\n", **q); }
Options- A. Error: cannot convert from void** to int**
- B. Garbage value
- C. 0
- D. No output Discuss
More questions
Correct Answer: True
Correct Answer: No
Correct Answer: 10.100000
Correct Answer: 2, 3, 3
Explanation:
Step 1: int x=4, y, z; here variable x, y, z are declared as an integer type and variable x is initialized to 4.
Step 2: y = --x; becomes y = 3; because (--x) is pre-decrement operator.
Step 3: z = x--; becomes z = 3;. In the next step variable x becomes 2, because (x--) is post-decrement operator.
Step 4: printf("%d, %d, %d\n", x, y, z); Hence it prints "2, 3, 3".
Step 2: y = --x; becomes y = 3; because (--x) is pre-decrement operator.
Step 3: z = x--; becomes z = 3;. In the next step variable x becomes 2, because (x--) is post-decrement operator.
Step 4: printf("%d, %d, %d\n", x, y, z); Hence it prints "2, 3, 3".
Correct Answer: No
Explanation:
A function can return floating point value.
Example:
#include <stdio.h>
float sub(float, float); /* Function prototype */
int main()
{
float a = 4.5, b = 3.2, c;
c = sub(a, b);
printf("c = %f\n", c);
return 0;
}
float sub(float a, float b)
{
return (a - b);
}
Output:
c = 1.300000
Correct Answer: c=1
Explanation:
Step 1: int a=100, b=200, c;
Step 2: c = (a == 100 || b > 200);
becomes c = (100 == 100 || 200 > 200);
becomes c = (TRUE || FALSE);
becomes c = (TRUE);(ie. c = 1)
Step 3: printf("c=%d\n", c); It prints the value of variable i=1
Hence the output of the program is '1'(one).
Step 2: c = (a == 100 || b > 200);
becomes c = (100 == 100 || 200 > 200);
becomes c = (TRUE || FALSE);
becomes c = (TRUE);(ie. c = 1)
Step 3: printf("c=%d\n", c); It prints the value of variable i=1
Hence the output of the program is '1'(one).
Correct Answer: 1, 40, 1
Explanation:
Step 1: int x=55; here variable x is declared as an integer type and initialized to '55'.
Step 2: printf("%d, %d, %d\n", x<=55, x=40, x>=10);
In printf the execution of expressions is from Right to Left.
here x>=10 returns TRUE hence it prints '1'.
x=40 here x is assigned to 40 Hence it prints '40'.
x<=55 returns TRUE. hence it prints '1'.
Step 3: Hence the output is "1, 40, 1".
Step 2: printf("%d, %d, %d\n", x<=55, x=40, x>=10);
In printf the execution of expressions is from Right to Left.
here x>=10 returns TRUE hence it prints '1'.
x=40 here x is assigned to 40 Hence it prints '40'.
x<=55 returns TRUE. hence it prints '1'.
Step 3: Hence the output is "1, 40, 1".
Correct Answer: BC
Correct Answer: k is a pointer to a pointer to a pointer to a pointer to a char
Correct Answer: 0
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