Discussion
Home ‣ C Programming ‣ Functions See What Others Are Saying!
- Question
What will be the output of the program?
#include<stdio.h> int i; int fun1(int); int fun2(int); int main() { extern int j; int i=3; fun1(i); printf("%d,", i); fun2(i); printf("%d", i); return 0; } int fun1(int j) { printf("%d,", ++j); return 0; } int fun2(int i) { printf("%d,", ++i); return 0; } int j=1;
Options- A. 3, 4, 4, 3
- B. 4, 3, 4, 3
- C. 3, 3, 4, 4
- D. 3, 4, 3, 4
- Correct Answer
- 4, 3, 4, 3
ExplanationStep 1: int i; The variable i is declared as an global and integer type.Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.
Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.
Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.
Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.
Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function.
Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function.
Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Hence the output is "4 3 4 3".
More questions
- 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
Correct Answer: Error: unknown max in declaration/Constant expression required
Explanation:
Step 1: A macro named MAX is defined with value 128Step 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.
- 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
Correct Answer: i= FFE4 ptr=10 j=FFE2 ptr=20
- 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
Correct Answer: Error: ++needs a value
Explanation:
This program will show an error "Cannot modify a const object".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.
- 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
Correct Answer: Error: cannot modify const object
- 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
Correct Answer: Error: cannot convert ptr const value
Explanation:
Step 1: char mybuf[] = "India"; The variable mybuff is declared as an array of characters and initialized with string "India".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".
- 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
Correct Answer: Error: cannot convert ptr const value
Explanation:
Step 1: char mybuf[] = "India"; The variable mybuff is declared as an array of characters and initialized with string "India".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".
- 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
Correct Answer: Hello
Explanation:
Step 1: const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string.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".
- 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
Correct Answer: 20
Explanation:
Step 1: int get(); This is the function prototype for the funtion get(), it tells the compiler returns an integer value and accept no parameters.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".
- 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
Correct Answer: -11, 34
Explanation:
Step 1: const c = -11; The constant variable 'c' is declared and initialized to value "-11".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
- 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
Correct Answer: Hi
Explanation:
switch(i) has the variable i it has the value '1'(one).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.
- 1. Point out the error in the program (in Turbo-C).