Adding to String array and reference - c++

Question

I have dynamic string's array:
string *acc = new string[2];
string some_string;
I would like to add two strings to this array, but there is problem with reference ( i think).
I have following example (very ugly, but it shows the problem):
for ( int i = 0; i < 2; i ++ ) {
if ( i == 0 )
some_string = "ab";
else
some_string = "cd";
acc[i] = some_string;
some_string = "";
}
return acc
Of course this code is without any sense, but my code is more complicated and it would hide the problem.
The point is that code is returning cd instead of abcd. In my opinion some_string = ""; make mess here. Am I right?
Is it possible to do it keeping logic of code?

Answer 1

I really don't think you should write code this way. And you don't need temporary some_string - assigning strings is well-defined and do exactly what it should. In thi case, use vector - it is much safer approach:
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main()
{
vector<string> ac;
for(int i = 0; i < 2; ++i)
{
if(i == 0)
ac.push_back("ab");
else
ac.push_back("cd");
}
cout<<ac[0]<<ac[1]<<endl;
return 0;
}
Working sample: here.

Answer 2

It is not clear why you expect "abcd" as result.
You assign some_string the value "cd" instead of adding to "ab".
Maybe this is what you are trying:
string* func()
{
string *acc = new string[2];
string some_string;
for ( int i = 0; i < 2; i ++ )
{
if ( i == 0 )
some_string = "ab"; // "ab"
else
some_string = some_string + "cd"; // "ab" + "cd" -> "abcd"
acc[i] = some_string;
//some_string = "";
}
return acc;
}
int main(int argc, char** argv)
{
string* y = func();
cout << y[0] << endl; // First element "ab"
cout << y[1] << endl; // Second element "abcd"
// Other stuff
// ....
// Remember to delete y
return(0);
}
Output:
ab
abcd
But as others had said.... Use vector instead.
You could also use std::array like this:
array<string, 2> func2()
{
array<string, 2> res;
string some_string;
for ( int i = 0; i < 2; i ++ )
{
if ( i == 0 )
some_string = "ab";
else
some_string = some_string + "cd";
res[i] = some_string;
}
// Or just this instead of a for loop
// res[0] = "ab";
// res[1] = "abcd";
return res;
}
int main(int argc, char** argv)
{
array<string, 2> z = func2();
cout << z[0] << endl;
cout << z[1] << endl;
return(0);
}

Answer 3

No, you simply made a bug:
for ( int i = 0; i < 2; i ++ ) {
if ( i == 0 ) // == instead of = !!!!!!!!!!!!!!!!!!!!!!
some_string = "ab";
else
some_string = "cd";
acc[i] = some_string;
some_string = "";
}
return acc;
Think over using a std::vector or std::list instead of pointer too. With a vector, your code looks like:
std::vector<std::string> vs;
...and later...
vs.push_back("ab");
vs.push_back("cd");