problem in understanding a part of a code

Asked Nov 23 '21 at 19:15

Active Nov 23 '21 at 19:29

Viewed 26 times

the code below is a part of a program which makes some integer sets by using class. the member variable is called set which is a boolean vector and set[i]=true if i is in set. now we want to have the members of a set in a string in the way that each member seprates by ','. for example if in main function we have IntegerSet A and the members of A are 1,2,3 ; by using toString.A() we should have {1,2,3} in output. I have the code of this part of program but my problem is that I don't know what bool first=true exactly do and how by using it the compiler understand that we mean what? I would appreciate if anyone can help me.

std::string IntegerSet::toString(IntegerSet)
{
    std::string str;
    bool first = true;
    str += '{';
    for (int i = 0; i <= limit; i++)
    {
        if (set[i])
        {
            if (!first)
                str += ",";
            else
                first = false; 
            str += std::to_string(i);
                
        }
    }
    str += '}';
    return str;
}

edited Nov 23 '21 at 19:29

πάντα ῥεῖ

asked Nov 23 '21 at 19:15

pantea

`bool first=true` assigns the value `true` to the boolean variable `first`. – Scott Hunter Nov 23 '21 at 19:18
`if (!first)` prevents adding a comma before the 1st value added to the string. – πάντα ῥεῖ Nov 23 '21 at 19:18
@ScottHunter What an on-point answer fo SO. – sweenish Nov 23 '21 at 19:18
@πάνταῥεῖ first is a separate part of the set. How does It understand if we are at the 1st value or not? – pantea Nov 23 '21 at 19:25
@pantea Because `first` is set to `true` for the 1st iteration of the loop, and it's not part of the set itself but a separate boolean variable. – πάντα ῥεῖ Nov 23 '21 at 19:27
@πάνταῥεῖ I can't understand the connection between `if(set[i])` and `if(!first)` in a loop – pantea Nov 23 '21 at 19:34
@pantea there's no connection actually. – πάντα ῥεῖ Nov 23 '21 at 19:35

problem in understanding a part of a code

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