hu************@yahoo.com said:
<snip>
>
1 #include <assert.h>
You don't use this, so you could just remove it.
By the way, the line numbers are a nuisance, since they prevent us from
easily compiling your code to syntax-check it.
2 #include <stdio.h>
This line brings in various prototypes and stuff for standard I/O. It's fine
- no problem with it.
What you are missing, though, is a prototype for toupper. I recommend that
you do this as well:
#include <ctype.h>
This will give you the necessary prototype for toupper.
3
4 char *StringUp(char *p)
This line declares that StringUp is a function that takes a pointer to a
character and returns a pointer to a character. It's fine - no problem
here.
5 {
This line begins the definition of the function body.
6 char *string_start=p;
This line defines a pointer to char, and initialises it with the value
passed via the parameter p.
7 while(*p=toupper(*p))++p;
This line needs a slight modification. Your function has no guarantee that
the characters it is processing are representable as unsigned char, so it
would be wiser to write the line as follows:
while(*p=toupper((unsigned char)*p))++p;
It would be wiser still to separate the loop body line from the controlling
line, as this makes the code a little easier to read:
while(*p=toupper((unsigned char)*p))
++p;
And it would be fantastically wise to use a compound statement (although it
is not, strictly speaking, required):
while(*p=toupper((unsigned char)*p))
{
++p;
}
because, in the general case, it eases the process of modifying the loop to
incorporate any extra statements that you might require at some later
point.
8 return string_start;
This is fine - you're returning the value you passed originally, which you
carefully copied into string_start at the beginning of your function.
9 }
This line closes the function definition.
10
11 int main(void)
This line declares main to be a function taking no parameters and returning
an int. No problem here.
12 {
13 char array[]="The man walked by the 3rd door and turned
right.";
C doesn't support the wrapping of string literals around a newline
character. I realise that your code probably doesn't have this problem -
it's just something that happened in your news client. Still, it makes
compiling the code at this end a bit of a nuisance. If you have very long
string literals, you can fold them like this:
char array[]="The man walked by the 3rd"
" door and turned right.";
Note that there is no ,,,comma,,, in between the two lines.
The C preprocessor takes this absence of non-whitespace separators as a sign
that we're talking about just one string literal, so it glues the pieces
together for you - it's just as if you'd written:
char array[]="The man walked by the 3rd door and turned right.";
14 printf("--%s\n",StringUp(array));
15 }
Ok, let's now say the new code is in lines 1-15 above (thanks Roland!).
If in line 8 I have "return string_start;", I get my desired results.
Is what is being returned a pointer?
Yes.
I thought pointers had to be prefaced with an "*". Yes?
When you declare them, yes. In use, though, it depends on what you want.
In your case, you want the pointer value itself, not the thing pointed to.
return string_start;
means "return the value this pointer has", which is precisely what you need.
return *string_start; /* no! */
would mean "return the value of the character pointed to by string_start",
which is not what you want.
Is there some way I can printf the memory addresses of the content of
*string_start?
*string_start is a single character.
string_start is a single pointer value (colloquially, an address).
Which do you mean?
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
