Using char on variables.
#10

I tested now the code below:
pawn Код:
new testchar[20 char] = {30, 1, 254, 105, 20};
    printf("%i", testchar[0]);
Yes, it's printed the number 30, but when I did print(testchar[0]) it didn't work... maybe cause I should use
the %i for the integers of the variable?

In addition, the thing I need to do is to *4 the number of the cells, right?
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Messages In This Thread
Using 'char' array on variables. - by Activest - 03.03.2013, 19:32
Re: Using char on variables. - by Misiur - 03.03.2013, 20:02
Re: Using char on variables. - by Vince - 03.03.2013, 20:02
Re: Using char on variables. - by ReVo_ - 03.03.2013, 20:03
Re: Using char on variables. - by AndreT - 03.03.2013, 20:05
Re: Using char on variables. - by Activest - 03.03.2013, 20:27
Re: Using char on variables. - by LarzI - 03.03.2013, 20:36
Re: Using char on variables. - by Vince - 03.03.2013, 20:36
Re: Using char on variables. - by LarzI - 03.03.2013, 20:44
Re: Using char on variables. - by Activest - 03.03.2013, 20:57

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