RE: [Dev-C++] compileresults

[Schedler, K. <Kur...@hi...>]

It means your function is expecting a pointer to a character (char *) and
you are sending it a character (char).  A char* holds the address of a
character.  What you need to do is call your function like this:
result(&mychar, myint).  The '&' means 'address of'.  Either that or change
the definition of your function to take a char instead of a char*.
"Casting" is forcing a variable of one type to be considered another type.
For example, if you had an integer pointer (int*) variable, ptr, you could
force it to be seen as a character pointer (char*) by 'casting' it like
this:  (char *)ptr.  Since you were passing a variable to a function that
was expecting a different type, the compiler assumed that you had forgotten
to cast your variable explicitly to the expected type.
 
-----Original Message-----
From: Ruth Aviton [mailto:av...@01...]
Sent: Friday, April 20, 2001 9:02 AM
To: dev...@li...
Subject: [Dev-C++] compileresults



 
hi everybody,
I am just a beginner in programming and I have two questions: 
1. When I compile a small, simple program i get a compile message as
follows. I have no idea what it means and have no idea where to start
looking!   passing `char' to argument 1 of `result(char *, int)' lacks a
cast This happens when i try to call a function.
2. Is there any place where i can get a explanation on what the compiler
results mean ? 
Thank you!
Ruth

 
 

			
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