returning vector<string> in c++

Question

I get: C:\Documents and Settings\ppp\Pulpit\Zadanie3Infix\main.cpp|72|error: conversion from 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >*' to non-scalar type 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >' requested|

and i don't know why. The function suppose to return vector of strings and it does, i used all of these * & and it does not work. Please correct me so it will return this vector.

it works like that s = "(2+23)" it will return it will return vector with tokens.

  vector<string> split(string s){
    vector<string>* v = new vector<string>();
    int i = 0; bool b;string el = "";
    while(i < s.size()){

        if(isNotNumber(s.at(i))){
            el = s.at(i);
            v->push_back(el);
            cout<<el<<" to operator"<<endl;
            el = "";
            i++;
        }else{
            while( i <s.size() && !isNotNumber(s.at(i))){//LENIWE WYLICZANIE !!!
                el = el + s.at(i);
                cout<<s.at(i)<<" to liczba"<<endl;
                i++;
            }
            v->push_back(el);
            el = "";
        }
    }
    cout<<"PO while"<<endl;
    for(int i = 0; i < v->size();i++){
        cout<<v->at(i)<<endl;
    }

    return v;
}

and what about that

stack<string>* stack = new stack<string>();

type specifier before stack ;////////////////

Change it to "vector<string>* split(string s){". you are returning pointer not by value — Arunmu
– Arunmu, Commented Aug 25, 2012 at 15:47
I would suggest doing the opposite of what ArunMu said. Keep the return value the same, and change the object in the function to match. — Benjamin Lindley
– Benjamin Lindley, Commented Aug 25, 2012 at 15:49
@ArunMu The OP really should pass a vector by reference. It's absurd to rely on pointers for this scenario. — chrisaycock
– chrisaycock, Commented Aug 25, 2012 at 15:54
@ArunMu: More costly? How so? Likely not in C++03 (rvo), and certainly not in C++11 (move semantics). — Benjamin Lindley
– Benjamin Lindley, Commented Aug 25, 2012 at 15:54
If you just want to make it work I suggest you take Benjamin Lindley's suggestion. Declare your vector as vector<string> v and return it by value. C++ is not Java so stop using new on everything. — Blastfurnace
– Blastfurnace, Commented Aug 25, 2012 at 16:01

chrisaycock · Accepted Answer · 2012-08-25 15:48:47Z

4

Your function split() claims to return vector<string>, however your return value is v. What is v? It's

vector<string>* v

You are attempting to return a pointer but your function declaration states that it is not returning a pointer. Actually, there's no need for a pointer at all in your function; just use a regular vector<string> or pass in a vector by reference:

void split(string s, vector<string>& v){

answered Aug 25, 2012 at 15:48

chrisaycock

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Comments

Jerry Coffin · Accepted Answer · 2012-08-25 16:47:21Z

3

The question about whether to return a pointer or a value, or pass in a reference to an existing vector was settled a long time ago: don't do any of the above. Instead, create a generic algorithm that works with iterators:

template <class InIt, class OutIt>
void tokenize(InIt b, InIt e, OutIt o) {
    while (b != e) {
        std::string token;
        while (isdigit(*b))
            token.push_back(*b++);
        if (!token.empty())
           *o++ = token;
        while (!isdigit(*b) && b != e)
            *o++ = std::string(1, *b++);
    }
}

To put the result in a vector of strings, you'd do something like:

std::string i("(2+23)");
std::vector<std::string> tokens;

tokenize(i.begin(), i.end(), std::back_inserter(tokens));

Or, if you wanted to display the tokens, one per line for testing:

tokenize(i.begin(), i.end(), std::ostream_iterator<std::string>(std::cout, "\n"));

...which will produce:

(
2
+
23
)

That fits with what I'd think of as the tokens in this input string. I should add, however, that while it works for that input, it may or may not produce what you expect for other inputs. In particular, it will treat whitespace characters as tokens, which you usually don't want (but does seem to be how the original was intended to work).

answered Aug 25, 2012 at 16:47

Jerry Coffin

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1 Comment

Benjamin Lindley Over a year ago

I would have to disagree that it's settled. Not everything has to be completely generic. Certainly you would agree that auto tokens = split("(2+23)"); is much more readable than your suggested method. That's gotta be worth something. I think both methods have valid reasons to exist, and perhaps one could make use of the other.

mathematician1975 · Accepted Answer · 2012-08-25 15:49:05Z

2

Your function signature returns a vector, but your return value is a pointer. Change the return value type to pointer to vector or return a vector by value in your function body.

answered Aug 25, 2012 at 15:49

mathematician1975

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Comments

Phillip Ngan · Accepted Answer · 2012-08-26 08:16:05Z

1

This is really Blastfurnace's answer, ...

but convert the line:

vector<string>* v = new vector<string>();

into

vector<string> v;

and change all the v-> into v.

This way you fix the memory leak too. (The memory allocated by new vector<string>() is never deleted). However, vector<string> v; puts the memory on the stack and is automatically deleted when it goes out of scope.

answered Aug 26, 2012 at 8:16

Phillip Ngan

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