String in C

I know there's really no strings in C. But I need some help if you may:

#include<stdio.h>

char password(char pass[]);

char password(char pass[])
{
return pass;
}

May I ask, if I want to return pass what data type do I need? If I use char it errors.

Thanks again,

string in C is really a char array.
You can see the function accept char[], so it should return the same.
In standard C, i think/guess that char* would be more correct.

If you work with strings I suggest you look up how pointers work, if you haven't already.

Yeah. You're right. I had to go more-in-depth about pointers.

I read in my book on C that the variable also takes a value as a pointer.

#include<stdio.h>

char pass(char msg[]);

char pass(char msg[])
{
return *msg;
}

main()
{
printf("%s",pass("Bob"));
getchar();
}

Why isn't this working? Keep in mind, my main scenerio I changed was *msg

OK, array names and pointers are mostly equivalent -- there are a few things that you can do with pointers that you can't do with arrays. So there shouldn't be any difference between pass(char msg[]) and pass(char *msg).

Where you're messing up is in the return type. If you want to return a char array, you're actually returning a char pointer. The way you have pass written now, you're only returning the first char of msg, which then gets misinterpreted in the printf since you're using "%s"; if you tried "%c" instead, you should see 'B' printed -- I have no idea what you're getting now.

This should work:

Note that I changed the return type to char* and I removed the * in the return statement that dereferenced msg. Also, I corrected the declaration of main and added the requisite return from main.

Look it over, think it over, and understand what's happening.

That worked. Thank you.

But I'm confused why you put a pointer by the function.

"*pass"

Why did you put a pointer there? I didn't see you use an address of operator so does the function have its own address?

Can you just please explain why you put a pointer there? That'll be great!

Purely stylistic.

char *pass
is the same as
char* pass
and even as
char*pass
It's not saying anything about being a pointer to pass, but rather just says that this is a char pointer datatype. It's most common to write it as
char *pass
even though some like to place it with the char to make that more clear.

I think the common way is due to multiple variable declarations on a single line; eg:
char ch, *cp;
That declares two variables: ch which is a char and cp which is a char pointer. As you can see, the * is associated with the variable name there, so that appears to be carried over to other declarations; eg, function parameter lists, function return types.

So to reiterate, there's no difference between char *pass and char* pass. They both say the exact same thing.

@swapy: hello

Blam! I just blew your program out of the water! Or as Da-Wei would put it, I sent it running off into the weeds to puke all over its own shoes.

I'm quite sure that we've already warned you about using gets. I also know that most documentation on gets warns you to not use it (I know that most man pages do) and that some compilers will issue a warning when you use it (or are you ignoring warnings?).

And I'm puzzled by your giving the array's dimension in the function header like that. Frankly, in more than 20 years experience with C/C++ I've not seen that done outside of a few people on this forum. I assume that it must work, but I don't see how it would make much difference.

Here is a much better version:


PS
My reply was primarily to address your implicit recommendation to use gets. Here is what the GNU C Library help file says:


When a buffer overflow occurs, the memory after the buffer get overwritten, AKA "clobbered". When this happens inside of a function (even main is a function), then that can destroy information that it vital to the execution of that function, including the address that it is to return to. This will most commonly lead to the program crashing.

But even worse, such careless programming leaves your program open to buffer overflow exploits, a common method for hackers to take over your system. Sure, the most malicious thing a hacker could do to your student program is to cause it to crash, but if you continue your careless practices in writing commercial software, then a hacker will find that buffer overflow and create an exploit tailored to it, one which will "return" to code embedded in the overflow input, the "payload", that will either take over or destroy that system. At best, it will allow the hacker to do a denial of service attack by causing the system to crash.

Learn as early as you can to program correctly and robustly.