Too Few Arguments

Code:

---

#include <iostream>
#include <stdio.h>

using namespace std;


char GetUserInput(void)

{
char Input;



    cout << "Input Residue Sequence: "<< endl;
    cin >> Input;
    return Input; //Returns a copy of Input
}   

char GetInputChange(char theInput)

{
 cout << "WOOORK"<< endl;
 cout << theInput << endl;
 cin.get();
 
 return (theInput);
         
}   

int main (int argc, char* argv[])

{
   
    GetUserInput ();
   
    GetInputChange();
 
cin.get();
}

---

Need some help, i don't know what results in getting the error, "To few Arguments for Function Char GetInputChange"

Just don't quite understand what the damn thing wants :)

Edit for mistake thnks :)

The error is probably Too few arguments for GetInputChange, not GetUserInput. The GetInputChange function takes a single argument, but when you call it in main you don't pass anything to it. When calling a function that takes an argument, you must pass it something. In this case, you probably want to store the char returned from GetUserInput and pass that variable to GetInputChange.

Hmm,

I'd hoped that the first function would just take user input, single char and store it in variable "Input", return Input keeping it for use to other functions?

and then the second function(GetInputChange) would be able to get what is in Input and just print it out ie (cout << theInput)

I thought placing "theInput" into the functions Header allowed it to be called for cout.

Am I going about this wrong?

Thanks for the help

Got it to compile thanks.

Did what i thought you ment, but i could be wrong.

Code:

---

int main (int argc, char* argv[])

{
   
    GetUserInput ();
   
    char Input = GetUserInput();
   
    GetInputChange(Input);
   
 {
        return Input;
 }
   
cin.get();
}

---

Will now move on to getting If/Else in now

close, but not quite

Code:

---

int main (int argc, char* argv[])

{
   
    GetUserInput ();
   
    char Input = GetUserInput();
   
    GetInputChange(Input);

// don't need these lines.
/*   
 {
        return Input;
 }
  */ 
cin.get();
}

---

Don't you need to put a return 0; at the end? :)
Since that won't even compile without some return statement... or shouldn't anyway.

slow down, you're not there yet - I assume that the below is what
your code looks like now - read comments:

Code:

---

#include <iostream>
#include <stdio.h>

//<stdio.h> is a C header - the C++ version is <cstdio>, but you
//don't need it in the program so there's no point in that include

using namespace std;


char GetUserInput(void)

{
char Input;



    cout << "Input Residue Sequence: "<< endl;
    cin >> Input;
    return Input; //Returns a copy of Input
}   

char GetInputChange(char theInput)

{
 cout << "WOOORK"<< endl;
 cout << theInput << endl;
 cin.get();
 
 return (theInput); // why are you returning the parameter you passed?
         
}   

//int main (void) would suffice - this version is to allow
//command line arguments - more complicated than your level
int main (int argc, char* argv[])

{
   
    GetUserInput (); //unnecessary call here - you throw away the return value here
   
    //and then calling it properly here
    char Input = GetUserInput();
   
    GetInputChange(Input);
   
 {//unnecessay braces here
        return Input; //Input is a char - main returns int
 }//and here
   
cin.get();
}

---

@dpro - Chaos is correct - compiler will automatically put in return
0 if you leave it out.

Just realised it didn't actaually print out "Wooork" :D

Thanks for the heads up

Thanks Both of you,

Quote:

---

return (theInput); // why are you returning the parameter you passed?

---

Sorry mistake from copying above functions return, don't quite know why i did it there.

Quote:

---

//int main (void) would suffice - this version is to allow
//command line arguments - more complicated than your level
int main (int argc, char* argv[])

---

Force of habit from this c++ book i read/copy from, everything else has it in so i add it to be sure.

Quote:

---

GetUserInput (); //unnecessary call here - you throw away the return value here

---

I thought it was needed to call function, wasn't sure (char Input = GetUserInput();) called it again, until i realised now why two
"Input Residue Sequence: " appeared :)

Quote:

---

return Input; //Input is a char - main returns int

---

To be Honest i had no idea return was based on the functions type sadly :(, ie Int for main.

Once again thanks all of you, a book just doesn't cut it in some areas, especially the original error.

Quote:

---

Originally Posted by Richie T

@dpro - Chaos is correct - compiler will automatically put in return
0 if you leave it out.

---

Ah, well thats good, but for someone who is new, its always good to give a full picture whether or not its needed. Personally, its a little nit, but whatever works I suppose.

Quote:

---

Originally Posted by dpro

Ah, well thats good, but for someone who is new, its always good to give a full picture whether or not its needed. Personally, its a little nit, but whatever works I suppose.

---

the other issue is that putting the "return input" in will skip the cin.get()

Code:

---

// original code
int main (int argc, char* argv[])
{
   
    GetUserInput ();
   
    char Input = GetUserInput();
   
    GetInputChange(Input);

 {
        return Input;
 }
// after the return statement!!
cin.get();
}

---

but yes, you're right. in general it is better to have a return 0 at the end of main.

Oh yes the return Input was in the wrong place, I meant at the end after cin.get(), but my point lacks any new information. So don't worry bout it :) Although technically it should be there, its not that big a deal.