Functions that return char[]

[ejjunju]

I have written this code as part of a project. if my variable Name has the value "Amazon", how can i change this function so that the output is say Amazon.BAS if x=1.?

char* read(int x)
{
string s3;

switch (x)
{
case 1: s3=".BAS";break;
case 2: s3=".PAR";break;
case 3: s3=".DAT";break;
case 4: s3=".OBS";break;
case 5: s3=".RCH";break;
default:cout<<"wrong argument in read(arg)\n";
};

//Make string s1 contain value held by variable Name;
string s1;
stringstream out;
out << this->Name;
s1 = out.str();


string aFileS =s1+s3;

//convert filename string to Char
char aFile[100];
memset( aFile, '\0', 100 );//sets all values in char-array Radfile to "\0"
aFileS.copy( aFile, aFileS.size() );
cout<<aFile<<endl;
return aFile;
}

[firedraco]

Use the string.append() function (should be pretty self-explanatory)

[Scipio]

This is one way to do it:

#include <iostream>
#include <string>

using namespace std;

string chancename (string oldname);
int main()
{
    string name;
    cout<<"Insert name please:\n";
    cin>>name;
    string newname=chancename(name);
    cout<<newname;
    cin.ignore();
    return 0;
}

string chancename(string oldname)
{
       int x;
       string newname;
       cout<<"Please enter value:\n";
       cin>>x;
       switch (x)
{
case 1: newname=oldname+".BAS";break;
case 2: newname=oldname+".PAR";break;
case 3: newname=oldname+".DAT";break;
case 4: newname=oldname+".OBS";break;
case 5: newname=oldname+".RCH";break;
default:cout<<"wrong argument in read(arg)\n";
}
return newname;
}

[firedraco]

And @scipio: Don't give away code for free thx

[Scipio]

He said he needed it for a project :). Sorry, i dont want to make peoples homework.