Hello,
On my program I have 9 char named c1,...c9.
During the program, I write: gtch=getchar() (I should give a number between 1 and 9)
How can I modifiy the "value" of the character c1,...9 according to what number I enter?
For example, c4= 'x'. Then, I enter the number 4, so gtch='4'. How can I transform c4 to 'y'.
PS: I'm a begginer on C...
Thank you !
> On my program I have 9 char named c1,...c9.
Rather than doing
c1, c2, c3 etc
you use an array instead.
This gets you 9 chars labelled c[0] to c[8];Code:char c[9];
So perhaps for an easier life in this case
This gets you 10 chars labelled c[0] to c[9];Code:char c[10];
> During the program, I write: gtch=getchar() (I should give a number between 1 and 9)
This will get you a character '1' to '9'
To transform a single digit character containing a number into an integer, all you do then is
gtch = gtch - '0';
You can then do
c[gtch] = 'y';
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.
Thank you for your quick answer !
I still have a small error:
Here's what I write (whithout details):
Code:char cases[10]={'x0','x1',...'x9'}; int main (void) ... char gtch; gtch=getchar(); gtch=gtch-'0'; cases[gtch]='y'; --> in this line I have an error: array suscrpit has type char , so when I run my program no 'y' is visible.
Thank you for your time !
Well getchar() actually returns an int, so you should really have
Also, things like 'x0' are not characters, since you have two chars between single quotes.Code:int gtch; gtch=getchar(); if ( gtch != EOF ) { gtch=gtch-'0'; cases[gtch]='y'; }
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 understand the problem, Thank you !
