On Monday, 27 July 2015 at 15:50:11 UTC, Alex wrote:
readf(" %s", &dieNumber);
What happens here is a bit tricky and trips up a lot of
programmers: readf leaves the end-of-line character in the
buffer, which readln then sees as meaning its job is done.
When you enter, say, 5, then press enter, the input looks like:
"5\n". (\n represents the newline aka end-of-line character).
readf " %s" first skips any spaces. There's none, so it doesn't
matter.
Then it reads in a number. So it takes the 5, leaving the rest of
the input behind.... so that \n from pressing enter is still
there.
readln reads everything it can until it hits a \n, picking up
where readf left off. So the result is it immediately sees the \n
that readf stopped on and it also stops.
An easy fix is to stick an extra readln() after the readf:
readf(" %s", &dieNumber);
readln(); // tell it to skip the rest of the line
// the rest of your code is unchanged
The next thing you'll have to face is actually going again when
they enter Y. (BTW be aware that this is case sensitive and
readln leaves the newline at the end of the input too! So
entering "y" won't work. And when you enter "Y"... the program
will see "Y\n", so it won't match "let's go again" either.)
But to actually go again, you'll probably want to read about
loops. I'll let you play with it a little from here though.
