functions declaration, pointers, cast, .... CONFUSING

Hi all,

I'm in the most confusing period of learning the C language. What's worse is that my book keeps writing exercises using commands and method that have not been taught yet! Anyhow, could someone please try to explain WHY and WHEN a function would be declared like the following:

char * compare ( char * , char *);

I don't understand the asterisk before the function name. This is the function header:

char * compare ( chat *str1, char *str2 )

and it returns the following

if ( x > y )
return ( str1);
else if ( x < y )
return (str2);

As long as I didn't have the asterisk before the function name, I got tonnes of errors and warnings.
could someone please explain?
Rhodium

Strings are dealt with in C as pointers. Look at your function like this:

Code:

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char* compare ( chat *str1, char *str2 )

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You're returning a char pointer. If you used char by itself you would be returning a single byte.

Quote:

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Originally posted by personification
Strings are dealt with in C as pointers.

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A string is not a pointer. It just happens that people often use pointers to point to the first element of a string. That way, they don't have to be concerned about the length when pointing to its location.

Basically what the function is doing in this case is returning a pointer to the first element of a string.

if ( x > y )
return ( str1);
else if ( x < y )
return (str2);

if x is greater than y then it returns the memory address of the first element of str1
else if x is less than y it returns the memory address of the first element of str2

Thanx a LOT personification,

That made a lot of things clearer; thanks....
just so you'd know, this is the entire code

#include <stdio.h>
#include <string.h>

char *a = "was--up?";
char *b = "world!?";
char *compare( char * , char * );

int main ( void )
{
char *out;
out = compare( a , b );

printf(" The longer string is %s.\n", out);

return 0 ;
}

char *compare ( char *str1 , char *str2 )
{
int x, y;

x = strlen(str1);
y = strlen(str2);

if ( x > y )
return (str1);
else if ( x < y )
return (str2);
}

I still get a warning saying:

warning: control reaches end of non - void function
the programme works fine but I want to know the reason of the warning;

PS: could someone explain what casting is?

Quote:

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Originally posted by Rhodium
warning: control reaches end of non - void function
the programme works fine but I want to know the reason of the warning;

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That's because it is possible that x and y are the SAME in which case the function doesn't return a value, so garbage data will be returned. You need to make another return statement after

Quote:

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Originally posted by Rhodium
PS: could someone explain what casting is?

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casting is when you temporarily want value to be looked at as a datatype which it really isn't. For instance:

let's say you had a pointer to an int and you only wanted to "grab" the first byte of it rather than the whole thing. A char is one byte in length, so you can typecast the pointer to a pointer to a char and then dereference. It will access just the first byte. Example:

Code:

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int a,
    *ptr = &a;

    char b = *((char*)ptr); // stores the first byte of 'a' to b

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>casting is when you temporarily want value to be looked at as a datatype which it really isn't

I've never really been fond of this "it lets you fake it" kind of explanation. A cast is an operator that performs a conversion.

Quote:

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6.5.4 Cast operators

• 4 Preceding an expression by a parenthesized type name converts the value of the expression to the named type. This construction is called a cast.85) A cast that specifies no conversion has no effect on the type or value of an expression.86)

• 85) A cast does not yield an lvalue. Thus, a cast to a qualified type has the same effect as a cast to the unqualified version of the type.
  
  86) If the value of the expression is represented with greater precision or range than required by the type named by the cast (6.3.1.8), then the cast specifies a conversion even if the type of the expression is the same as the named type.

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Many times we encounter trivial conversions. Often this is a cast from a pointer to one object type to a pointer to another object type. An example might be converting an int* to a char*. But we should keep in mind that a "conversion" is performed nonetheless. And there are other times when the conversion is not trivial.

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While not wanting to dredge up the Casting malloc thread, there were some related comments that still bother me.

Quote:

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In the first place, there is nothing wrong with using type coercion. newbies call this 'casting'.

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Wow! Newbies wrote the C standard!

Quote:

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Type coercsion was developed for one reason and one reason only--To resolve type mismatches at the preprocessor level. type coercion has nothing to do with execution.

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This is simply not correct. An example might be casting a floating point value to a fixed point value. On one system, this conversion was implemented using a call to an assembly routine at runtime. So it had everything to do with execution.

Quote:

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I will say it simpler so you understand. "type coercion" or "casting" is used solely to make the compiler see two different types of variable as the same thing, so long as they are the same size.

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Again, this is not correct. The sizes do not need to be the same. Ever convert a char to an int? A long to a short? Between floating and fixed point (again)? Explicity or implicitly, casts (conversions) are done all the time. Do your best to be aware of this. And "used solely...", uh, no.

Quote:

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In the first place, there is nothing wrong with using type coercion. newbies call this 'casting'

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Who the hell are you quoting? No one in this thread said that. None of those quotes are from this thread.

Quzah.

He said those quotes were from the casting malloc thread.:p