### Discussion :: Variable Number of Arguments - Find Output of Program (Q.No.3)

Pooja said: (Aug 5, 2010) | |

Please describe how to solve this problem. |

Raj said: (Jul 30, 2011) | |

First dumplist function is called with elements 2, 4 and 8. When va_start is called, it points to element 2. When it sees va_arg, it points to the next element 4. Since va_arg is within while loop, it prints 4 8 for the first function and 6 9 7 for the second function. So the answer is option C. |

Prince Bhanwra said: (Oct 21, 2011) | |

what does while loop condition indicates?? i.e. while(n-->0) what this condition indicates???? |

Anil said: (Jan 3, 2012) | |

while(n-- >0) ...loop until the value of n is greater than 0. I think here n represents the no of arguments passed excluding n. |

Niyati said: (Feb 18, 2012) | |

Please can anyone clearly explain this problem.I am not able to understand this.please.. |

Anon said: (Aug 19, 2012) | |

@Raj... When it sees va_arg, it points to the next element 4. Since va_arg is within while loop, it prints 4 8 for the first function and 6 9 7 for the second function.What you are saying here is incorrect.va_arg returns the element it is pointing to and the move to next element in list |

Vimal Dahduk said: (Nov 12, 2012) | |

@Niyati. while(n-- > 0) means u understand so n is indicate total number of argument. va_start is indicate first argument. va_args(p,int) return first argument value and va_args(p,int) second argument value. so that print first function and second function and not print first value. Print first value by simply write printf("%d", n); of the above example. I hope you understand. |

Harini said: (Mar 22, 2014) | |

Actually n-->0 means that, n-- > 0 i.e. n is first decremented and is compared with 0. |

Prince Bansal said: (Apr 17, 2014) | |

Wants to know more. Read this code. #include<stdio.h> //Header file contains Macros #include<stdarg.h> //Compute avg of given no's double avg(int num, ...) { int i; double sum=0.0; //Create va_list variable va_list valist; //Initializa va_list with num va_start(valist,num); //Run untill all variables are visted once for(i=1;i<=num;++i) { //va_arg will pick up variable from argument list and each //time move 4 bytes ahead (because we are using here integer //second argument in va_arg. sum+=va_arg(valist,int); } return sum/num; } int main() { printf("\nAVG:%e",avg(5, 1,2,3,4)); return 0; } |

K.karthik Raja said: (Jul 1, 2014) | |

Please clearly explain this problem? |

Rahul said: (Jul 11, 2014) | |

Can anyone please explain this in detail. @Raj well, ques2 also talks about the same concept, but here, va_arg doesn't seem to point to the next argument. |

Priya said: (Sep 19, 2015) | |

I can't understand, please anyone explain the above program clearly. |

Kalpana said: (Sep 25, 2015) | |

I need a clear explanation about this program. |

Anonymous said: (Oct 1, 2015) | |

Here first variable passed in both calls will be considered as the starting point to work with variable list. In first part, va start : 2. So n would be, 2 so loop runs for 4 and 7. Whereas in second case, va start:3 so 6 9 and 7 is printed on the screen. |

Vinayk Patil said: (Nov 7, 2015) | |

In both dumplist 2, 3 initial values represent no of arguments taken remaining are values stored so n->0 indicates for 1st while (n->0) indicates n is equal to 0 for both times of dumplist. It is checked but initial value of n=2 no of arguments taken by dumplist accordingly n=3 for 2nd one n--is done according to printf function respective values included initially is printed. |

Vvvvvvv said: (Jul 21, 2016) | |

In the list, the first parameter always represents the number of arguments to be passed. So the 2 in the first function represents that there are 2 arguments. And 3 in second function represents there are 3 arguments. So first it prints 4 and 8. Next, it will print 6, 9, 7. Excluding 4 and 8. 4 and 8 are actually to define a number of parameters. And they are not get printed. |

Sandipan said: (Oct 6, 2016) | |

What is va_end? |

Teju said: (Mar 30, 2017) | |

while(n-->0) Explain this line. |

Mahi said: (Jun 17, 2017) | |

Decrement value of n. Untill (n--) to 0. |

Akshaya Shelke said: (Jul 7, 2017) | |

Thanks for your explanation @Vvvvvvv. |

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