Discussion
Home ‣ C Programming ‣ Declarations and Initializations See What Others Are Saying!
- Question
What is the output of the program in Turbo C (in DOS 16-bit OS)?
#include<stdio.h> int main() { char *s1; char far *s2; char huge *s3; printf("%d, %d, %d\n", sizeof(s1), sizeof(s2), sizeof(s3)); return 0; }
Options- A. 2, 4, 6
- B. 4, 4, 2
- C. 2, 4, 4
- D. 2, 2, 2
- Correct Answer
- 2, 4, 4
ExplanationAny pointer size is 2 bytes. (only 16-bit offset)
So, char *s1 = 2 bytes.
So, char far *s2; = 4 bytes.
So, char huge *s3; = 4 bytes.
A far, huge pointer has two parts: a 16-bit segment value and a 16-bit offset value.Since C is a compiler dependent language, it may give different output in other platforms. The above program works fine in Windows (TurboC), but error in Linux (GCC Compiler).
More questions
- 1. 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".
- 2. 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- 3. 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).- 4. 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".- 5. 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
- 6. 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
- 7. 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
- 8. What will be the output of the program?
#include<stdio.h> #include<string.h> int main() { static char s[] = "Hello!"; printf("%d\n", *(s+strlen(s))); return 0; }
Options- A. 8
- B. 0
- C. 16
- D. Error Discuss
Correct Answer: 0
- 9. What will be the output of the program?
#include<stdio.h> int main() { int i=4, j=8; printf("%d, %d, %d\n", i|j&j|i, i|j&&j|i, i^j); return 0; }
Options- A. 4, 8, 0
- B. 1, 2, 1
- C. 12, 1, 12
- D. 0, 0, 0 Discuss
Correct Answer: 12, 1, 12
- 10. What will you do to treat the constant 3.14 as a float?
Options- A. use float(3.14f)
- B. use 3.14f
- C. use f(3.14)
- D. use (f)(3.14) Discuss
Correct Answer: use 3.14f
Explanation:
Given 3.14 is a double constant.
To specify 3.14 as float, we have to add f to the 3.14. (i.e 3.14f)
Comments
There are no comments.
More in C Programming:
Programming
Copyright ©CuriousTab. All rights reserved.
- 1. What will be the output of the program?