C++ :- Replacing a piece of string by another string

Question

I am trying to write a REPLACE function that will replace the given string by the requiredstring. When I am dry running the function on paper, everything seems to be fine but while executing, it's not giving the correct output. The code is as follows:-

string REPLACE(string src,string reqd,string given)
{
    int i,j,k;
    int pos = FIND(src,given);
    if(pos==-1)
        return "";
    else
    {
        char *arr = new char[src.length()+reqd.length()-given.length()];  // creating the array that will hold the modified string
        for(i=0;i<pos;i++)
            arr[i] = src[i];     // copying the initial part of the string
        for(i=pos,j=0;i<pos+reqd.length()+1&&j<reqd.length();i++,j++)
            arr[i] = reqd[j];    // copying the required string into array
        for(i=pos+reqd.length()+1,k=0;i<sizeof(arr);i++,k++)
            arr[i] = src[pos+given.length()+k];   // copying the remaining part of source string into the array
        return arr;
    }   
}

Here the FIND is also written by me and has been tested in many cases. I don't see any error in FIND.

Why are you manually allocating memory? Isn't std:string good enough? You also leak memory every time you call this function. — jrok
– jrok, Commented Sep 2, 2013 at 15:22
But then, I cannot make the string flexible enough to handle modifications that change their lengths. — kusur
– kusur, Commented Sep 2, 2013 at 15:27
Yes you can, use std::string arr(src.length()+reqd.length()-given.length(), 0);. — john
– john, Commented Sep 2, 2013 at 15:28
Sure you can,std::string is a dynamic container that can change its size. — jrok
– jrok, Commented Sep 2, 2013 at 15:29

jrok · Accepted Answer · 2013-09-02 15:39:18Z

1

for(i=pos+reqd.length()+1,k=0; i<sizeof(arr); i++,k++)
//                               ^^^^^^^^^^^
//                           This is always the same

sizeof(arr) is the same as sizeof(char*), a compile-time constant value. You need to keep the size of a dynamically allocated array yourself (or, better yet, simply use std::string instead).

answered Sep 2, 2013 at 15:39

jrok

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1 Comment

kusur Over a year ago

Works fine if I replace sizeof(arr) by src.length()+reqd.length()-given.length()+1

Abhishek Bansal · Accepted Answer · 2013-09-02 15:29:32Z

1

I don't think it is a good idea to mix std::string with char arrays. Following should work:

string REPLACE(string src,string reqd,string given)
{
    int pos = FIND(src,given);

    src.replace( pos, given.size(), reqd );
    return src;    
}

answered Sep 2, 2013 at 15:29

Abhishek Bansal

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2 Comments

kusur Over a year ago

Thanks for the advice. Could you please elaborate on why using char arrays and string together is a bad idea?

Abhishek Bansal Over a year ago

The class std::string itself can be used like a char array if required. Other than that, as already shown, it provides a lot more functionality and clarity to your code. It also automatically takes care of dynamic memory allocation without the fear of memory leaks.

sara · Accepted Answer · 2013-09-02 15:25:27Z

0

Assuming you do not want to re-invent the wheel:

string REPLACE(string src,string reqd,string given)
{
    string str(src);
    size_t pos = str.find(given);
    if(pos == std::string::npos)
        return "";
    str.replace(pos, given.length(), reqd);
    return str;
}

answered Sep 2, 2013 at 15:25

sara

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