CODEWITHSUNDEEP
c++searchstl
ubbdd
ubbdd

Reputation: 241

return type for std::search()

I am trying to test an example of search() from Stroustup's book.

string quote("why waste time learning, when ignorance is instantaneous?");

bool in_quote(const string& s){
    char* p = search(quote.begin(), quote.end(), s.begin(), s.end());
    return p != quote.end();
}

void test(){
    bool b1 = in_quote("learning"); // b1=true
    bool b2 = in_quote("lemming"); // b2=false
}

But I get the following error:

error C2440: 'initializing' : cannot convert from
'std::_String_iterator<_Elem,_Traits,_Alloc>' to 'char *'

It looks like the return type is not right. I also tried string::iterator, and got the same error. So, what should be the right type, is it supposed to be the iterator type of the container? Thanks

Upvotes: 3

Views: 708

Answers (5)

Mark Ransom
Mark Ransom

Reputation: 308402

Early implementations of string could easily have used char* as their iterator type, allowing this incorrect code snippet to compile properly. Most modern implementations of string::iterator have a proper class type and would not be convertible to char*.

The signature for std::search is:

template <class ForwardIterator1, class ForwardIterator2>
   ForwardIterator1 search ( ForwardIterator1 first1, ForwardIterator1 last1,
                             ForwardIterator2 first2, ForwardIterator2 last2 );

As you can see, the return type is the same as the type of the first two iterators passed to the function. In your case string::iterator should have worked, unless there is some part of the code you did not show us that made quote const in which case you could use string::const_iterator.

Upvotes: 1

Gnawme
Gnawme

Reputation: 2399

Per SGI documentation, the form of search you are using has the signature:

template <class ForwardIterator1, class ForwardIterator2>
ForwardIterator1 search(ForwardIterator1 first1, ForwardIterator1 last1,
                        ForwardIterator2 first2, ForwardIterator2 last2);

Since your FowardIterator1 type is std::string::iterator, your return type must be std::string::iterator as well.

Upvotes: 0

fredoverflow
fredoverflow

Reputation: 263220

How about simply not caring about the return type? :)

bool in_quote(const string& s){
    return search(quote.begin(), quote.end(), s.begin(), s.end()) != quote.end();
}

Upvotes: 4

MartinStettner
MartinStettner

Reputation: 29174

I tried the following

bool in_quote(const string& s){
  string::iterator p = search(quote.begin(), quote.end(), s.begin(), s.end());
  return p != quote.end();
}

And it did compile without error ...

Upvotes: 3

Yakov Galka
Yakov Galka

Reputation: 72519

You have a const string, so it must be a const_iterator:

string::const_iterator p = search(quote.begin(), quote.end(), s.begin(), s.end());

Upvotes: 1

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