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- Question
What do the following declaration signify?
void *cmp();
Options- A. cmp is a pointer to an void type.
- B. cmp is a void type pointer variable.
- C. cmp is a function that return a void pointer.
- D. cmp function returns nothing.
- Correct Answer
- cmp is a function that return a void pointer.
- 1. What do the following declaration signify?
char *scr;
Options- A. scr is a pointer to pointer variable.
- B. scr is a function pointer.
- C. scr is a pointer to char.
- D. scr is a member of function pointer. Discuss
- 2. Declare the following statement?
"A pointer to a function which receives nothing and returns nothing".
Options- A.
void *(ptr)*int; - B.
void *(*ptr)() - C.
void *(*ptr)(*) - D.
void (*ptr)()
Discuss
- 3. What do the following declaration signify?
char **argv;
Options- A. argv is a pointer to pointer.
- B. argv is a pointer to a char pointer.
- C. argv is a function pointer.
- D. argv is a member of function pointer. Discuss
- 4. Declare the following statement?
"A pointer to a function which receives an int pointer and returns float pointer".
Options- A.
float *(ptr)*int; - B.
float *(*ptr)(int) - C.
float *(*ptr)(int*) - D.
float (*ptr)(int)
Discuss
- 5. What do the following declaration signify?
int *f();
Options- A. f is a pointer variable of function type.
- B. f is a function returning pointer to an int.
- C. f is a function pointer.
- D. f is a simple declaration of pointer variable. Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
Complicated Declarations problems
Search Results
Correct Answer: scr is a pointer to char.
Correct Answer:
void (*ptr)()Correct Answer: argv is a pointer to a char pointer.
Correct Answer:
float *(*ptr)(int*)Correct Answer: f is a function returning pointer to an int.
Correct Answer:
char (*ptr)[3];Correct Answer: cmp is a pointer to a function which returns void .
Correct Answer: ptr is a array of 30 pointers to integers.
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.
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.
Correct Answer: 0, 2, 1
Explanation:
enum takes the format like {0,1,2..) so pass=0, fail=1, atkt=2
stud1 = pass (value is 0)
stud2 = atkt (value is 2)
stud3 = fail (value is 1)
Hence it prints 0, 2, 1
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