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- Question
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
- Correct Answer
- Error: cannot convert ptr const value
ExplanationStep 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".
Constants problems
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- 1. Point out the error in the program.
#include<stdio.h> int main() { const int k=7; int *const q=&k; printf("%d", *q); return 0; }
Options- A. Error: RValue required
- B. Error: Lvalue required
- C. Error: cannot convert from 'const int *' to 'int *const'
- D. No error Discuss
Correct Answer: No error
Explanation:
No error. This will produce 7 as output.- 2. Point out the error in the program.
#include<stdio.h> int main() { const int x; x=128; printf("%d\n", x); return 0; }
Options- A. Error: unknown data type const int
- B. Error: const variable have been initialised when declared.
- C. Error: stack overflow in x
- D. No error Discuss
Correct Answer: Error: const variable have been initialised when declared.
Explanation:
A const variable has to be initialized when it is declared. later assigning the value to the const variable will result in an error "Cannot modify the const object".Hence Option B is correct
- 3. 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".
- 4. Point out the error in the program.
#include<stdio.h> const char *fun(); int main() { *fun() = 'A'; return 0; } const char *fun() { return "Hello"; }
Options- A. Error: RValue required
- B. Error: Lvalue required
- C. Error: fun() returns a pointer const character which cannot be modified
- D. No error Discuss
Correct Answer: Error: fun() returns a pointer const character which cannot be modified
- 5. 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.
- 6. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> fun(...); int main() { fun(3, 7, -11.2, 0.66); return 0; } fun(...) { va_list ptr; int num; va_start(ptr, n); num = va_arg(ptr, int); printf("%d", num); }
Options- A. Error: fun() needs return type
- B. Error: ptr Lvalue required
- C. Error: Invalid declaration of fun(...)
- D. No error Discuss
Correct Answer: Error: Invalid declaration of fun(...)
Explanation:
There is no fixed argument in the definition fun()- 7. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> int main() { void display(int num, ...); display(4, 12.5, 13.5, 14.5, 44.3); return 0; } void display(int num, ...) { float c; int j; va_list ptr; va_start(ptr, num); for(j=1; j<=num; j++) { c = va_arg(ptr, float); printf("%f", c); } }
Options- A. Error: invalid va_list declaration
- B. Error: var c data type mismatch
- C. No error
- D. No error and Nothing will print Discuss
Correct Answer: Error: var c data type mismatch
Explanation:
Use double instead of float in float c;- 8. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> int main() { void display(char *s, int num1, int num2, ...); display("Hello", 4, 2, 12.5, 13.5, 14.5, 44.0); return 0; } void display(char *s, int num1, int num2, ...) { double c; char s; va_list ptr; va_start(ptr, s); c = va_arg(ptr, double); printf("%f", c); }
Options- A. Error: invalid arguments in function display()
- B. Error: too many parameters
- C. Error: in va_start(ptr, s);
- D. No error Discuss
Correct Answer: Error: in va_start(ptr, s);
Explanation:
We should have use va_start(ptr, num2);- 9. Point out the error if any in the following program (Turbo C).
#include<stdio.h> #include<stdarg.h> void display(int num, ...); int main() { display(4, 'A', 'a', 'b', 'c'); return 0; } void display(int num, ...) { char c; int j; va_list ptr; va_start(ptr, num); for(j=1; j<=num; j++) { c = va_arg(ptr, char); printf("%c", c); } }
Options- A. Error: unknown variable ptr
- B. Error: Lvalue required for parameter
- C. No error and print A a b c
- D. No error and print 4 A a b c Discuss
Correct Answer: No error and print A a b c
- 10. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> void varfun(int n, ...); int main() { varfun(3, 7, -11, 0); return 0; } void varfun(int n, ...) { va_list ptr; int num; num = va_arg(ptr, int); printf("%d", num); }
Options- A. Error: ptr has to be set at begining
- B. Error: ptr must be type of va_list
- C. Error: invalid access to list member
- D. No error Discuss
Correct Answer: Error: ptr has to be set at begining
Explanation:
Using va_start(ptr, int);
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- 1. Point out the error in the program.