Working with strings using pointers

Question

So hello guys, its my code:

#include <iostream>
#include <string>

using namespace std;

void writeDown(string*t)
{
    for (int i = 0; *(t+i)!=NULL; i++)
    {
        cout << *(t+i) <<endl;
    }
}

int main()
{
    string t;
    getline(cin, t);
    string *wsk = &t;
    writeDown(wsk);
    return 0;
}

So I simply insert a string, and program should cout<< every single char from it in a new line. But here is what pops out:

binary '!=' : no operator found which takes a left-hand operand of type 'std::string' (or there is no acceptable conversion)

What am I doing wrong?

Btw. I am working in VS 2013 for Win Desktop (btw. vol.2 - is it a good environment for C++ coding? - got 2015, but it's to slow on my laptop)

It looks like you're trying to treat std::string* like it's a char*. That's not how it works, they are very different things. — Captain Obvlious
– Captain Obvlious, Commented Jan 11, 2016 at 2:46
If std::string is analogous to const char*, to what would std::string* be analogous? — erip
– erip, Commented Jan 11, 2016 at 2:51
@Frynio if my answer worked for you, will you be kind enough to mark it as accepted? — Nacho
– Nacho, Commented Feb 8, 2016 at 16:33

Nacho · Accepted Answer · 2016-01-11 03:08:22Z

2

Well that code is bad in so many levels, but just to keep it simple and answer your question I would probably use iterators. Forget about that old C style if you are using C++.

Try this:

void writeDown(const string & t)
{
    for (string::const_iterator it = t.begin();
        it != t.end();
        ++it)
    {
        cout << *it << endl;
    }
}

Note that writeDown no longer takes a pointer as parameter but a constant reference to the string you want to print.

You will have to change your main to call writeDown:

string t;
getline(cin, t);
writeDown(t);

If you insist on using the "array" method you would have to use it like this, but ...

void writeDown(const string & t)
{
    for (int i = 0; t.c_str()[i] != NULL; i++)
    {
        cout << t.c_str()[i] << endl;
    }
}

answered Jan 11, 2016 at 3:08

Nacho

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3 Comments

minecraftplayer1234 Over a year ago

Is it the same thing (by "the same" I mean speed) when I use asterix & or * as a parameter?

Captain Obvlious Over a year ago

@Frynio Premature optimization is the root of stupid. Concentrate on writing code that's easy to understand and maintain then optimize based on the results of profiling.

minecraftplayer1234 Over a year ago

Thank you. I would always do some sort of unimportant things, focusing on them, not on my main goal

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