How Backwards Functions

I have been rambling through different threads on the cprogramming.com borad, and have noticed the function that follows.

Code:

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Mid(s1, 3, 0) = "123";

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Now I don't know the proper name for this type of function, but from what I read from the thread was that this finction takes the "123" and adds it at position 3 to string s1. Now I would like to make a function similar to the one above, but my c++ book does not cover this type/style of function so I have no idea of how to make this type of function. I would look it up on the internet, but I don't know what to call this type of function. If someone would either give me a referance, or at least told me what to look under I would much apriciate it.

-JLBShecky

Look up pointers and references, functions: passing by value and passing by reference, classes, constructors, and operator overloading. With about 1 month of studying, you should be able to understand them.

Judging by the code, you're talking about the "Duplicate VB Mid" function contest. You can download the contestants code and have a look through it, that might give you some ideas.

Some words to look up:
assignment operator
functor (or function object)
Also, check out this link: http://www.parashift.com/c++-faq-lite/references.html

Yes I was talking about the "Duplicate VB Mid" contest thread. I now about passing by reference and by value, I also know abit about class, constructors, and operator overloading, but as of yet I do not under stand how to get a function to work on the left side of the assinement operator "=". I tried looking for the contestants code for the "Duplicate VB Mid" contest, but I could not find any it. Thanks for the current input.

-JLBShecky

I don't know anything about the mid function, but my guess is that this line:

Code:

---

Mid(s1, 3, 0) = "123";

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first of all is a constructor, and second it is assigning "123" to s1 (maybe means string 1?)

Mid(s1, 3, 0) = "123";

Here's a class designed so you can use that format. It overloads the operator() and the operator=:

Code:

---

#include <iostream>
#include <cstring>

using namespace std;

class A
{
public:
        A(char* f, char* t, int n);
        A();

        A& operator()(char* str, int b, int c);
        A& operator=(char* str);
       
        void display()
        {
                cout<<"flag = "<<flag<<endl;
                cout<<"text = "<<text<<endl;
                cout<<"number = "<<number<<endl;
        }

        ~A()
        {
                delete [] flag;
                delete [] text;
        }
       
private:
        char* flag;
        char* text;

        int number;

};

A::A(char* f, char* t, int n)
{
        flag = new char[strlen(f) + 1];
        strcpy(flag, f);
       
        text = new char[strlen(t) + 1];
        strcpy(text, t);

        number = n;
}

A::A()
{
        flag = new char[4];
        strcpy(flag, "off");

        text = new char[6];
        strcpy(text, "empty");
       
        number = 1;
}

A& A::operator()(char* str, int b, int c)
{
        number += b;
        number *= c + 1;

        delete [] flag;
        flag = new char[strlen(str) + 1];       
        strcpy(flag, str);

        return *this;
}

A& A::operator=(char* str)
{

        delete [] text;
        text = new char[strlen(str) + 1];
        strcpy(text, str);
        return *this;
}




int main()
{
        A Mid("off", "blue", 5);
        char* s1 = "on";

               
        Mid(s1, 3, 0) = "123";



        Mid.display();                       
        return 0;
}

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I don't see how we would be able to justify a class for this function since it is only really a function in VB and not an object. The simplest approach would be to create a function that takes four arguments instead of three.

Something like this.

Mid(s1, 3, 0, "123");

Further more the result must exist in s1 and not in an object.

>I don't see how we would be able to justify a class for this function since it is only really a function in VB and not an object.
I agree that simplicity is best, but C++ gives us the power to do fun things in our spare time as well. :)

>but as of yet I do not under stand how to get a function to work on the left side of the assinement operator "=".
You can do it by returning a reference to an object. Unfortunately, strings are much harder to work with this manner, which is probably why there was a contest for it.

Here's my take on this issue (using the contest's rules and testing procedures), extensive error checking has been omitted:

Code:

---

#include <iostream>
#include <string>

class StrRef {
    bool dynamic;
    std::string *base;
    int index;
    int overwrite;
public:
    StrRef(const char *init, int at, int rep);
    StrRef(std::string& init, int at, int rep);
    ~StrRef();
     
    std::string operator=(std::string s);
    std::string operator=(StrRef ref);
     
    operator std::string();
};

StrRef::StrRef(const char *init, int at, int rep):
    dynamic(true), base(new std::string(init)),
    index(at), overwrite(rep)
{}

StrRef::StrRef(std::string& init, int at, int rep):
    dynamic(false), base(&init),
    index(at), overwrite(rep)
{}

StrRef::~StrRef()
{
    if (dynamic)
        delete base;
}

std::string StrRef::operator=(std::string s)
{
    base->replace(index, overwrite, "");
    base->insert(index, s);
    return *base;
}

std::string StrRef::operator=(StrRef ref)
{
    base->replace(index, overwrite, "");
    base->insert(index, *ref.base, ref.index, ref.base->length());
    return *base;
}

StrRef::operator std::string()
{
    return base->substr(index, overwrite);
}

StrRef Mid(const char *s, int at, int rep = 0)
{
    return StrRef(s, at, rep);
}

StrRef Mid(std::string& s, int at, int rep = 0)
{
    return StrRef(s, at, rep);
}

// Contest test program
int main()
{
    using namespace std;

    bool correct = true;
    string s1,s2; 
   
    //Basic test
    s2 = "0123456789";
    string s3 = Mid(s2,1,4);
    s1 = Mid(s2,1,4);
    s2 = Mid("0123456789",1,4);
    if (s1 != "1234" || s2 != "1234" || s3 != "1234")
        cout << "Basic test incorrect : " << s1, correct=false;
   
   
    //First test
    s1 = "abcdefg";
    Mid(s1, 3, 0) = "123";
    if (s1 != "abc123defg")
        cout << "First test incorrect : " << s1, correct=false;
   
   
    //Second test
    s1 = "abcdefg";
    s2 = "123";
    Mid(s1, 3, s2.size()) = s2;
    if (s1 != "abc123g")
        cout << "Second test incorrect : " << s1, correct=false;
   
    //Third test
    s1 = "abc";
    s2 = "123";
    Mid(s1, 1, 1) = Mid(s2,1);
    if (s1 != "a23c")
        cout << "Third test incorrect : " << s1, correct=false;   
   
    if (correct)
        cout << "All tests correct!";

    cin.get();
}

---

Tested on the following compilers:

G++ 3.2 (no warnings or errors, correct output)
VC++ 6 (no warnings or errors, correct output)
DevC++ 4.9.8 (no warnings or errors, correct output)
Borland C++ 6 (no warnings or errors, correct output)
Borland C++ 5.5 (no warnings or errors, correct output)

Prelude,

I'm trying to follow your code. For the required functionality:

std::string mystring = Mid(someotherstring, 2, 5);

it seems to me that Mid() needs to return a string type or a char* because operator= for the string class has no idea what a StrRef object is, but your Mid() function returns a StrRef object. I think this is the key bit of trickery:

operator std::string();

but I don't understand the format of that function. No return type? But, in the function definition you appear to return a string object in your return statement:

StrRef::operator std::string()
{
return base->substr(index, overwrite);
}

I think I'm the guy in your avator right now.

operator std::string() allows implicit conversion from class StringRef to std::string, that means that you can use a StringRef in place of an std::string without casting.

For example, if you had a function with the following signature:
void DoSomething(std::string& str);
you'd be able to pass a StringRef object, as the compiler would do the conversion internally by calling operator std::string().

Quote:

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Originally posted by zMan
I don't see how we would be able to justify a class for this function since it is only really a function in VB and not an object. The simplest approach would be to create a function that takes four arguments instead of three.

Something like this.

Mid(s1, 3, 0, "123");

---

So? The rules in VB don't carry over to C++. The objective of the subjective contest was to try and duplicate the functionality and syntax of this VB "function", not to devise the simplest approach to substring replacement.

Ok, thanks. Now I understand what it does, but I still don't understand the format. Do you know of any links I could go to?

Is the operator an old style cast? Hmmm...that must be it. What about not having a return type, and then returning a std::string in the return statement? How come you get implicit functionality and aren't required to explicitly make the cast?

edit: Sheesh! Never mind, it's in my C++ book.

Due everyone's input I now understand how the works "Mid(s1, 3, 0) = "123";". My main problem was I was just thinking of a plain function and not of using a class. I gust have to start to think more object orientantedly. I also want to say thanks to all the input.

-JLBShecky