Simple program on Arrays
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.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
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.
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.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
Thanks !
I got it.
> I use gcc in Ubuntu OS
And did it complain about void main?
Or do you just ignore all warnings?
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
> I use gcc in Ubuntu OS
And did it complain about void main?
Or do you just ignore all warnings?
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
