Copy part of a string directly to vector error: no match operator=

Question

I want to implement a simple function that will split std::string based on the delimiter ,\s.

For example, split("Emerson, Lake, Palmer") should yield std::vector<std::string> of

Emerson
Lake
Palmer

I discovered, that it could be simply achieved with back_inserter() (the example is very helpful) . The function below iterates over supplied string and copies its content directly to a vector.

std::vector<std::string> split(const std::string& s) {
    std::vector<std::string> tokens;

    std::string::const_iterator from = s.begin();
    for (std::string::const_iterator it = s.begin(); it < s.end(); it++) {
        if ( *it == ',' ) {
            std::copy(from, it, std::back_inserter(tokens));
            from = it + 2;
        }
        if (it == s.end()-1)
            std::copy(from, it+1, std::back_inserter(tokens));
    }
    return tokens;
}

However, the code doesn't compile with the error

no match for 'operator=' (operand types are 'std::back_insert_iterator<std::vector<std::__cxx11::basic_string<char> > >' and 'const char')

I roughly understand what does it mean, however, I don't know how to go about this. Am I missing something simple to make it work, or the whole concept is not a way to go?

`tokens` is a `vector` so you can add strings: `tokens.emplace_back(std::string(from, it));` — 001, Jan 18 '21 at 22:05
This isn't a good use of `back_inserter`. Pair `back_inserter` with a function from that eats the input stream as an iterator and stores into a container iterator. — user4581301, Jan 18 '21 at 22:08
@Johny's answer gets the job done, however, I am curious what's wrong with this approach. @user4581301, I thought `std::copy` is such a function. — Maverick, Jan 18 '21 at 22:12
@JohnnyMopp passing temporary to `emplace_back()` defeats whole purpose of such functions. you basically make it the same as `push_back()` — Slava, Jan 18 '21 at 22:12
@Maverick yes `std::copy()` such a function but to use it here you need to iterate over something that has `std::sting` inside ie `*it` for iterator provides string, not individual symbol. — Slava, Jan 18 '21 at 22:15
You know, you've got me there, @Maverick. I was looking at the question wrong. I knew you were using `std::copy` to gather characters into a string and that the back inserter was expecting a `string`, but the disjoint types totally ed up my brain. — user4581301, Jan 18 '21 at 22:19

score 2 · Accepted Answer · answered Jan 18 '21 at 22:07

2

When you submit a range as 2 iterators to std::copy() you mean that you want to copy each element that such iterators point to to the target. But std::string::iterator points to individual chars and I do not think you want to copy each symbol into the vector separately. Instead you can create a std::string from pair and submit it to the vector:

for (std::string::const_iterator it = s.begin(); it < s.end(); it++) {
    if ( *it == ',' ) {
        tokens.emplace_back( from, it );
        from = it + 2;
    }
...

answered Jan 18 '21 at 22:07

Slava

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Isn't it the approach you criticized in the comments under the question? ;> Nevertheless, maybe there is just nothing better when one wants to stick to my initial idea. Thanks for the help! – Maverick Jan 18 '21 at 22:27
What approach do you mean that I criticized? Please elaborate. – Slava Jan 18 '21 at 22:29
that passing temporary to emplace_back() defeats the whole purpose of such functions. – Maverick Jan 18 '21 at 22:31
1

If you look carefully you will see the difference and you should notice that I do not pass temporary of `std::string` but submit 2 iterators directly. That the whole purpose of `emplace...` functions - they forward arguments to ctor. You can find details here https://stackoverflow.com/questions/50740699/copy-elision-when-creating-object-inside-emplace – Slava Jan 18 '21 at 22:36
Indeed, I overlooked that one. All is clear now. Have a good day – Maverick Jan 18 '21 at 22:39

score 1 · Answer 2 · answered Jan 18 '21 at 22:09

In this statement

std::copy(from, it, std::back_inserter(tokens));

the iterator from supplies objects of the type char while the iterator of the type std::back_inserter expects an object of the type std::string.

Instead you could write at least

tokens.push_back( std::string( from, it ) );

or

tokens.push_back( { from, it } );

Copy part of a string directly to vector error: no match operator=

2 Answers2