Discussion
Home ‣ C Programming ‣ Control Instructions See What Others Are Saying!
- Question
What will be the output of the program?
#include<stdio.h> int main() { char j=1; while(j < 5) { printf("%d, ", j); j = j+1; } printf("\n"); return 0; }
Options- A. 1 2 3 ... 127
- B. 1 2 3 ... 255
- C. 1 2 3 ... 127 128 0 1 2 3 ... infinite times
- D. 1, 2, 3, 4
- Correct Answer
- 1, 2, 3, 4
- 1. Point out the error in the program (in Turbo-C).
#include<stdio.h> #define MAX 128 int main() { const int max=128; char array[max]; char string[MAX]; array[0] = string[0] = 'A'; printf("%c %c\n", array[0], string[0]); return 0; }
Options- A. Error: unknown max in declaration/Constant expression required
- B. Error: invalid array string
- C. None of above
- D. No error. It prints A A Discuss
- 2. What will be the output of the program in TurboC?
#include<stdio.h> int fun(int **ptr); int main() { int i=10, j=20; const int *ptr = &i; printf(" i = %5X", ptr); printf(" ptr = %d", *ptr); ptr = &j; printf(" j = %5X", ptr); printf(" ptr = %d", *ptr); return 0; }
Options- A. i= FFE2 ptr=12 j=FFE4 ptr=24
- B. i= FFE4 ptr=10 j=FFE2 ptr=20
- C. i= FFE0 ptr=20 j=FFE1 ptr=30
- D. Garbage value Discuss
- 3. What will be the output of the program?
#include<stdio.h> int main() { const int i=0; printf("%d\n", i++); return 0; }
Options- A. 10
- B. 11
- C. No output
- D. Error: ++needs a value Discuss
- 4. Point out the error in the program.
#include<stdio.h> #include<stdlib.h> union employee { char name[15]; int age; float salary; }; const union employee e1; int main() { strcpy(e1.name, "K"); printf("%s", e1.name); e1.age=85; printf("%d", e1.age); printf("%f", e1.salary); return 0; }
Options- A. Error: RValue required
- B. Error: cannot modify const object
- C. Error: LValue required in strcpy
- D. No error Discuss
- 5. Point out the error in the program.
#include<stdio.h> #define MAX 128 int main() { char mybuf[] = "India"; char yourbuf[] = "CURIOUSTAB"; char *const ptr = mybuf; *ptr = 'a'; ptr = yourbuf; return 0; }
Options- A. Error: unknown pointer conversion
- B. Error: cannot convert ptr const value
- C. No error
- D. None of above Discuss
- 6. Point out the error in the program.
#include<stdio.h> #define MAX 128 int main() { char mybuf[] = "India"; char yourbuf[] = "CURIOUSTAB"; char const *ptr = mybuf; *ptr = 'a'; ptr = yourbuf; return 0; }
Options- A. Error: cannot convert ptr const value
- B. Error: unknown pointer conversion
- C. No error
- D. None of above Discuss
- 7. What will be the output of the program?
#include<stdio.h> int main() { const char *s = ""; char str[] = "Hello"; s = str; while(*s) printf("%c", *s++); return 0; }
Options- A. Error
- B. H
- C. Hello
- D. Hel Discuss
- 8. What will be the output of the program?
#include<stdio.h> int get(); int main() { const int x = get(); printf("%d", x); return 0; } int get() { return 20; }
Options- A. Garbage value
- B. Error
- C. 20
- D. 0 Discuss
- 9. What will be the output of the program?
#include<stdio.h> int main() { const c = -11; const int d = 34; printf("%d, %d\n", c, d); return 0; }
Options- A. Error
- B. -11, 34
- C. 11, 34
- D. None of these Discuss
- 10. What will be the output of the program?
#include<stdio.h> int main() { int i = 1; switch(i) { printf("Hello\n"); case 1: printf("Hi\n"); break; case 2: printf("\nBye\n"); break; } return 0; }
Options- A. Hello
Hi - B. Hello
Bye - C. Hi
- D. Bye Discuss
More questions
Correct Answer: Error: unknown max in declaration/Constant expression required
Explanation:
Step 2: const int max=128; The constant variable max is declared as an integer data type and it is initialized with value 128.
Step 3: char array[max]; This statement reports an error "constant expression required". Because, we cannot use variable to define the size of array.
To avoid this error, we have to declare the size of an array as static. Eg. char array[10]; or use macro char array[MAX];
Note: The above program will print A A as output in Unix platform.
Correct Answer: i= FFE4 ptr=10 j=FFE2 ptr=20
Correct Answer: Error: ++needs a value
Explanation:
Step 1: const int i=0; The constant variable 'i' is declared as an integer and initialized with value of '0'(zero).
Step 2: printf("%d\n", i++); Here the variable 'i' is increemented by 1(one). This will create an error "Cannot modify a const object".
Because, we cannot modify a const variable.
Correct Answer: Error: cannot modify const object
Correct Answer: Error: cannot convert ptr const value
Explanation:
Step 2: char yourbuf[] = "CURIOUSTAB"; The variable yourbuf is declared as an array of characters and initialized with string "CURIOUSTAB".
Step 3: char *const ptr = mybuf; Here, ptr is a constant pointer, which points at a char.
The value at which ptr it points is not a constant; it will not be an error to modify the pointed character; There will be an error only to modify the pointer itself.
Step 4: *ptr = 'a'; The value of ptr is assigned to 'a'.
Step 5: ptr = yourbuf; Here, we are changing the pointer itself, this will result in the error "cannot modify a const object".
Correct Answer: Error: cannot convert ptr const value
Explanation:
Step 2: char yourbuf[] = "CURIOUSTAB"; The variable yourbuf is declared as an array of characters and initialized with string "CURIOUSTAB".
Step 3: char const *ptr = mybuf; Here, ptr is a constant pointer, which points at a char.
The value at which ptr it points is a constant; it will be an error to modify the pointed character; There will not be any error to modify the pointer itself.
Step 4: *ptr = 'a'; Here, we are changing the value of ptr, this will result in the error "cannot modify a const object".
Correct Answer: Hello
Explanation:
Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".
Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello".
Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s.
Hence the output of the program is "Hello".
Correct Answer: 20
Explanation:
Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20".
The function get() returns the value "20".
Step 3: printf("%d", x); It prints the value of the variable x.
Hence the output of the program is "20".
Correct Answer: -11, 34
Explanation:
Step 2: const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'.
Step 3: printf("%d, %d\n", c, d); The value of the variable 'c' and 'd' are printed.
Hence the output of the program is -11, 34
Correct Answer: Hi
Explanation:
Then case 1: statements got executed. so, it prints "Hi". The break; statement make the program to be exited from switch-case statement.
switch-case do not execute any statements outside these blocks case and default
Hence the output is "Hi".
Comments
There are no comments.More in C Programming:
Programming
Copyright ©CuriousTab. All rights reserved.