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- Question
What is the output of the program?
#include<stdio.h> int main() { extern int a; printf("%d\n", a); return 0; } int a=20;
Options- A. 20
- B. 0
- C. Garbage Value
- D. Error
- Correct Answer
- 20
Explanationextern int a; indicates that the variable a is defined elsewhere, usually in a separate source code module.printf("%d\n", a); it prints the value of local variable int a = 20. Because, whenever there is a conflict between local variable and global variable, local variable gets the highest priority. So it prints 20.
Declarations and Initializations problems
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- 1. What is the output of the program given below?
#include<stdio.h> int main() { enum status { pass, fail, atkt}; enum status stud1, stud2, stud3; stud1 = pass; stud2 = atkt; stud3 = fail; printf("%d, %d, %d\n", stud1, stud2, stud3); return 0; }
Options- A. 0, 1, 2
- B. 1, 2, 3
- C. 0, 2, 1
- D. 1, 3, 2 Discuss
Correct Answer: 0, 2, 1
Explanation:
enum takes the format like {0,1,2..) so pass=0, fail=1, atkt=2stud1 = pass (value is 0)
stud2 = atkt (value is 2)
stud3 = fail (value is 1)
Hence it prints 0, 2, 1
- 2. What will be the output of the program in 16 bit platform (Turbo C under DOS)?
#include<stdio.h> int main() { extern int i; i = 20; printf("%d\n", sizeof(i)); return 0; }
Options- A. 2
- B. 4
- C. vary from compiler
- D. Linker Error : Undefined symbol 'i' Discuss
Correct Answer: Linker Error : Undefined symbol 'i'
Explanation:
Linker Error : Undefined symbol 'i'
The statement extern int i specifies to the compiler that the memory for 'i' is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name 'i' is available in any other program with memory space allocated for it. Hence a linker error has occurred.- 3. What do the following declaration signify?
int *ptr[30];
Options- A. ptr is a pointer to an array of 30 integer pointers.
- B. ptr is a array of 30 pointers to integers.
- C. ptr is a array of 30 integer pointers.
- D. ptr is a array 30 pointers. Discuss
Correct Answer: ptr is a array of 30 pointers to integers.
- 4. What do the following declaration signify?
void (*cmp)();
Options- A. cmp is a pointer to an void function type.
- B. cmp is a void type pointer function.
- C. cmp is a function that return a void pointer.
- D. cmp is a pointer to a function which returns void . Discuss
Correct Answer: cmp is a pointer to a function which returns void .
- 5. Declare the following statement?
"A pointer to an array of three chars".
Options- A.
char *ptr[3](); - B.
char (*ptr)*[3]; - C.
char (*ptr[3])(); - D.
char (*ptr)[3];
Discuss
Correct Answer:char (*ptr)[3];
- 6. What will be the output of the program?
#include<stdio.h> int X=40; int main() { int X=20; printf("%d\n", X); return 0; }
Options- A. 20
- B. 40
- C. Error
- D. No Output Discuss
Correct Answer: 20
Explanation:
Whenever there is conflict between a local variable and global variable, the local variable gets priority.- 7. What is the output of the program
#include<stdio.h> int main() { extern int fun(float); int a; a = fun(3.14); printf("%d\n", a); return 0; } int fun(int aa) { return (int)++aa; }
Options- A. 3
- B. 3.14
- C. 0
- D. 4
- E. Compile Error Discuss
Correct Answer: Compile Error
Explanation:
2 Errors
1. Type mismatch in redeclaration of fun
2. Type mismatch in parameter aa- 8. In the following program how long will the for loop get executed?
#include<stdio.h> int main() { int i=5; for(;scanf("%s", &i); printf("%d\n", i)); return 0; }
Options- A. The for loop would not get executed at all
- B. The for loop would get executed only once
- C. The for loop would get executed 5 times
- D. The for loop would get executed infinite times Discuss
Correct Answer: The for loop would get executed infinite times
Explanation:
During the for loop execution scanf() ask input and then printf() prints that given input. This process will be continued repeatedly because, scanf() returns the number of input given, the condition is always true(user gives a input means it reurns '1').Hence this for loop would get executed infinite times.
- 9. What is the output of the program
#include<stdio.h> int main() { int a[5] = {2, 3}; printf("%d, %d, %d\n", a[2], a[3], a[4]); return 0; }
Options- A. Garbage Values
- B. 2, 3, 3
- C. 3, 2, 2
- D. 0, 0, 0 Discuss
Correct Answer: 0, 0, 0
Explanation:
When an automatic array is partially initialized, the remaining elements are initialized to 0.- 10. 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 Discuss
Correct Answer: 2, 4, 4
Explanation:
Any 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).
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- 1. What is the output of the program given below?