stl for_each return type

[rocketman03]

Hi,

I was trying to write a function to concat all the elements in a vector<string> into single string. I was trying to use the stl for_each algorithm and using its return type as the concatenated string. I was told that for_each has the same return type as the function provided to it, so why does this not compile ? and how do I get it to work ?

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

using namespace std;

string concat(string& s);
	
int main()
{
	vector<string> u(10, "Hello");
	string v;
	v.reserve(1024);
	v = for_each(u.begin(),u.end(),concat);
	cout << v << endl;
	return 0;
}
string concat(string& s)
{
	static string str;
	str += s;
	return str;
}

Thank you!

[anon]

Because it returns the pointer to the function concat, not a string.

It seems to be an abuse of the for_each algorithm, seeing that there is a special-purpose accumulate in the <numeric> header.

v = std::accumulate(u.begin(),u.end(),std::string());

For string manipulations you might also look into boost's string algorithms.

v = boost::join(u, "");

[rocketman03]

Ah, thank you! that seems like a much easier way to do it anyway.

Because it returns the pointer to the function concat, not a string.

Yes, I just found that out myself too, but could you give me an example of what you could use this return type for ?

[anon]

Yes, I just found that out myself too, but could you give me an example of what you could use this return type for ?

You could obtain the result of concatenation from for_each, if you used a function object instead. For example, like this:

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

using namespace std;

struct concat
{
    std::string result;
    void operator()(const std::string& s)
    {
        result += s;
    }
};

int main()
{
	vector<string> u(10, "Hello");
	string v;
	v.reserve(1024);
	v = std::for_each(u.begin(), u.end(), concat()).result;
	cout << v << endl;
	return 0;
}

[rocketman03]

Ahh, I tested it out and it worked, but I'm a little confused now.
the function in the struct which is being called from for_each is the overloaded operator which has a return type of void. But in your example it is as if an object of the struct is being returned.
Also, you are directly calling concat() without an object of the struct, and result is not static. So how does it retain its value ?

[anon]

Indeed an instance of the struct is returned and then the result member is accessed to get the value. And the call of concat() creates an instance of the struct.

A function object overloads operator(), so it can be used as if it was a function. But it is still an object that can have member variables.

[rocketman03]

Ahh, I just looked up it up. I believe this is the same thing as a functor ? But, it makes sense now, Thank you!

[Elysia]

It is a functor.