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Home ‣ C Programming ‣ Variable Number of Arguments Comments
- Question
While defining a variable argument list function we drop the ellipsis(...)?
Options- A. Yes
- B. No
- Correct Answer
- Yes
- 1. Can we pass a variable argument list to a function at run-time?
Options- A. Yes
- B. No Discuss
- 2. It is necessary to call the macro va_end if va_start is called in the function.
Options- A. Yes
- B. No Discuss
- 3. In a function that receives variable number of arguments the fixed arguments passed to the function can be at the end of argument list.
Options- A. True
- B. False Discuss
- 4. For a function receives variable number of arguments it is necessary that the function should receive at least one fixed argument.
Options- A. True
- B. False Discuss
- 5. A function that receives variable number of arguments should use va_arg() to extract arguments from the variable argument list.
Options- A. True
- B. False Discuss
- 6. Can we write a function that takes a variable argument list and passes the list to another function?
Options- A. Yes
- B. No Discuss
- 7. Can the fixed arguments passed to the function that accepts variable argument list, occur at the end?
Options- A. Yes
- B. No Discuss
- 8. Is it necessary that in a function which accepts variable argument list there should be at least be one fixed argument?
Options- A. Yes
- B. No Discuss
- 9. The macro va_arg is used to extract an argument from the fixed micro argument list and advance the pointer to the next argument.
Options- A. Yes
- B. No Discuss
- 10. What will be the output of the program?
#include<stdio.h> #include<string.h> int main() { char dest[] = {97, 97, 0}; char src[] = "aaa"; int i; if((i = memcmp(dest, src, 2))==0) printf("Got it"); else printf("Missed"); return 0; }
Options- A. Missed
- B. Got it
- C. Error in memcmp statement
- D. None of above Discuss
Variable Number of Arguments problems
Search Results
Correct Answer: No
Explanation:
Every actual argument list must be known at compile time. In that sense it is not truly a variable argument list.
Correct Answer: Yes
Correct Answer: False
Correct Answer: True
Correct Answer: True
Correct Answer: Yes
Correct Answer: No
Correct Answer: Yes
Correct Answer: No
Correct Answer: Got it
Explanation:
memcmp compares the first 2 bytes of the blocks dest and src as unsigned chars. So, the ASCII value of 97 is 'a'.
if((i = memcmp(dest, src, 2))==0) When comparing the array dest and src as unsigned chars, the first 2 bytes are same in both variables.so memcmp returns '0'.
Then, the if(0=0) condition is satisfied. Hence the output is "Got it".
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