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This is the basic array program which takes 2 array elements as input and gives that elements as output.PHP Code:// testing arrays
# include <stdio.h>
void main()
{
int a[2],i;
printf("\nenter elements of array A");
for (i=0;i<=1;i++)
{
printf("\nenter element =");
scanf("%d",&a[i]);
}
printf("\nthe entered array is %d ",a[2]);
}
I don't understand why my compiler is showing output as 2 no matter what you enter.
I use gcc in Ubuntu OS
Your array named a only has 2 elements, so a[2] does not exist. You should be printing the elements of a one by one.
1.can you give me the program code of it.
2. When I declared array for only 2 elements, It should print 2 elements if I asked it, why It should matter.
Yes, but no, I won't. Considering what you have already done, this should be a piece of cake.Quote:
Originally Posted by suryak
Indeed, you are able to print 2 elements. The problem is that you are not printing two elements. You print a single element a[2], but a[2] does not exist. a[0] and a[1] exist.Quote:
Originally Posted by suryak
Thanks !
I got it.
> I use gcc in Ubuntu OS
And did it complain about void main?
Or do you just ignore all warnings?
> I use gcc in Ubuntu OS
And did it complain about void main?
Or do you just ignore all warnings?