HI,
I wrote a test function to experiment with C's variable argument list (from #include<stdarg.h>) to accept varying number of inputs:
#include<iostream.h>
#include<stdarg.h>
#include<stdio.h>
dounble Average(int num, ...);
int main(int argc, char*argv[]){
int ave=Average(3, 12, 22, 4);
printf("Average = %d\n", ave);
return 0;
}
double Average(int num, ...){
va_list arg;
va_start(arg, num);
printf("\ n Number = %d\n", num);
int sum=0;
for (int i=0; i<num; i++){
printf("\n Sum b4 = %d\n", sum);
sum+=va_arg(arg, int);
printf("\n Sum b4 = %d\n", sum);
}
va_end(arg);
return sum/num;
}
Output:
Sum b4=0
Sum = 12
Sum b4 =12
Sum = 34
Sum b4 = 34
Sum = 38
Average =12
I was wondering if anyone could show me how I could access the elements in va_list? I'd like to print out the values in va_list. From the output, I'm not sure why it does not compute the number 3 into the average as well.
Can someone show me how it's done?
Thanks
This Question has been solved and asker verified All Experts Exchange premium technology solutions are available to subscription members.
Experts Exchange has been collecting answers to technology questions since 1996…3 million and counting! If you have a question, chances are we already have your answer.
If you can't find the exact answer you're looking for, ask our exclusive community of 50,000 experts. You’ll get a personalized answer from a trusted professional.
Thousands of free tech tips, tricks, how-to’s and tutorials are available in our peer reviewed articles section. See for yourself how smart our experts are, no login required.
Save solutions to your questions, answers you’ve discovered through searching plus helpful articles in your personal knowledgebase for easy future access.
Access the answers to your technology questions today.
30-day free trial. Register in 60 seconds.
Members of the expert community talk about why the experience at Experts Exchange is different than what you will find anywhere else.
Try it out and discover for yourself.
30-day free trial. Register in 60 seconds.
Join the community of experts here and help other tech pros by answering question in your area of expertise. You can earn FREE access to all Experts Exchange's premium features and resources.
This question didn't show any activity for more than 21 days. I will ask Community Support to close it unless you finalize it yourself within 7 days.
You can always request to keep this question open. But remember, experts can only help if you provide feedback to their comments.
Unless there is objection or further activity, I will suggest to accept
"Krap"
comment(s) as an answer.
PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER!
========
Werner
Business Accounts
Answer for Membership
by: KrapPosted on 2002-11-14 at 08:13:28ID: 7448908
This is due to the fact that "3" isn't part of the variable part of your arguments. It's a fixed parameter.
va_start(arg,num) means "start reading variable arguments in the arg variable, starting at the adress AFTER num's address".
This way, you can pass fixed AND variable arguments to your function, accessing fixed params with their name, and the variables one with va_arg, reading them in the order and knowing their size. You should allways initialize your variable list, with va_start, using the last fixed param.
For example, the printf(const char *format,...) function use the format string to know the size of the variable arguments (%s means it's a char *, %d an int, ...).
A really incomplete implementation should look like :
printf(const char *format, ...)
{
va_list arg;
char *cp;
int i;
long l;
float f;
// ...
char c1, c2;
int pos=0;
va_start(arg,format);
while (c1=format[pos++])
{
if (c1=='%')
{
if (c2=format[pos++])
{
switch (c2)
{
case 's' : cp=va_arg(arg,char *); break;
case 'd' : i=va_arg(arg,int); break;
case 'f' : f=va_arg(arg,float); break;
// ...
}
}
else
// ERROR
}
}
va_end(arg);
}
Hope this helps you to understand !