converting an array of null terminated const char* strings to a std::vector< std::string >

Question

I have a Visual Studio 2008 C++ function where I'm given an array of null-terminated strings const char* and a count of the number of strings in that array.

I'm looking for a clever way of turning an array of const char* in to a std::vector< std::string >

/// @param count - number of strings in the array
/// @param array - array of null-terminated strings
/// @return - a vector of stl strings
std::vector< std::string > Convert( int count, const char* array[] );

Boost is fine, STL is fine.

That's not an array of const char*. Do you mean that the strings are stored sequentially and separated by NULLs? — Ben Voigt
– Ben Voigt, Commented Dec 16, 2010 at 18:38
There, I fixed it since the OP wasn't responding and the question doesn't make any sense otherwise. — Edward Strange
– Edward Strange, Commented Dec 16, 2010 at 18:45
Of course it made sense, just a different from one you are suggesting — Gene Bushuyev
– Gene Bushuyev, Commented Dec 16, 2010 at 18:55
@Gene: The OP's code didn't match his question, since he said "I'm looking for a clever way of turning an array of const char* in to a std::vector<std::string>" (emphasis mine). He then used a single const char* as the function parameter. The question made far more sense than the code -- ergo the code was edited by Noah to reflect the question. To the extent that the OP meant the code instead, it's incumbent on him to say so -- not on you to downvote all the well-meaning answers to what was probably his question. — Stuart Golodetz
– Stuart Golodetz, Commented Dec 16, 2010 at 19:08

genpfault · Accepted Answer · 2010-12-16 18:42:42Z

15

Something like this?:

vector< string > ret( array, array + count );

answered Dec 16, 2010 at 18:42

genpfault

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7 Comments

Loki Astari Over a year ago

It always amazes me the poeple that try and do it the hard-way. +1 for keeping it simple.

Edward Strange Over a year ago

Of course, the great thing about this answer is that it is NOT clever.

PaulH Over a year ago

@Noah - You're right - I should have said 'simple' not 'clever'.

Stuart Golodetz Over a year ago

The other great thing about this answer is that it's one line long :)

Edward Strange Over a year ago

meh... One liners don't impress me unless they're truly the simplest method to perform the required task.

|

Loki Astari · Accepted Answer · 2010-12-16 21:16:08Z

4

Try this:

#include <string>
#include <vector>
#include <iterator>
#include <algorithm>
#include <iostream>


int main(int argc,char* argv[])
{
    // Put it into a vector
    std::vector<std::string>    data(argv, argv + argc);

    // Print the vector to std::cout
    std::copy(data.begin(), data.end(), std::ostream_iterator<std::string>(std::cout, "\n"));
}

edited Dec 16, 2010 at 21:16

answered Dec 16, 2010 at 20:11

Loki Astari

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Comments

Stuart Golodetz · Accepted Answer · 2010-12-16 18:41:39Z

3

Assuming that the signature of your function is inadvertently wrong, do you mean something like this?

#include <algorithm>
#include <iterator>
#include <string>
#include <vector>

std::vector<std::string> Convert(int count, const char **arr)
{
    std::vector<std::string> vec;
    vec.reserve(count);
    std::copy(arr, arr+count, std::back_inserter(vec));
    return vec;
}

int main()
{
    const char *arr[3] = {"Blah", "Wibble", "Shrug"};
    std::vector<std::string> vec = Convert(3, arr);
    return 0;
}

answered Dec 16, 2010 at 18:41

Stuart Golodetz

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2 Comments

PaulH Over a year ago

This is what I started with and why I decided to ask the question. I knew there was an easier way... My brain was just not bringing it to the surface.

Stuart Golodetz Over a year ago

@PaulH: Yup, it's very easy to forget that vector has a constructor designed for exactly this situation -- as I think I've proved in this case :)

Gene Bushuyev · Accepted Answer · 2010-12-16 19:02:46Z

0

I assume in your array strings are separated by 0s.

std::vector< std::string > Convert( int count, const char* array )
{
   std::vector< std::string > result;

   for(const char* begin = array; count; --count)
   {
    const char* end = begin;
    for(; *end; ++end);
    result.push_back(std::string(begin, end));
    begin = end + 1;
   }

   return result;
}

edited Dec 16, 2010 at 19:02

answered Dec 16, 2010 at 18:53

Gene Bushuyev

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2 Comments

Chris Hopman Over a year ago

This code works for the original question, sort of. That is, originally the questioner's function had this signature and so the assumption that the strings are separated by \0 is reasonable. In that case, this code works. However, the question was edited.

Edward Strange Over a year ago

Obviously not, since it was the old version I was commenting ON. But whatever.

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