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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 check if a bit in number is set or not.
Options- A. True
- B. False Discuss
Correct Answer: True
- 2. A pointer union CANNOT be created
Options- A. Yes
- B. No Discuss
Correct Answer: No
- 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
Correct Answer: 10.100000
- 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
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".
- 5. Functions cannot return a floating point number
Options- A. Yes
- B. No Discuss
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- 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
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).- 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
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".- 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
Correct Answer: BC
- 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
Correct Answer: k is a pointer to a pointer to a pointer to a pointer to a char
- 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
Correct Answer: 0
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