I'm trying to copy a part of a std::string into another string, but I cannot find any function that does what I want in the reference. Example: const int Length = 10; std::string temp = "abc def ghi jkl mno pqr stu vxyz"; std::string message = (temp.end(), temp.end()-Length); I want to pick out the last part of the original string, in this case the last 10 characters.

If you want to use iterators:

std::string message( temp.end()-Length, temp.end() );

Notes here: We're using std::string's 2 parameter constructor (which you can't call using assignment construction syntax... hence the lack of an '=' here). Also, standard C++ convention dictates that the beginning iterator is always first -- since temp.end()-Length is before temp.end(), the parameter order needs to be swapped.

If you want to accomplish the same thing with an existing std::string you can use it's assign function (or a temporary):



As for how your code is being handled: You're using the comma operator, which means your code behaves something like this:

temp.end();std::string message = temp.end()-Length;

Which is to say, it doesn't compile, since you can't assign a std::string to a single iterator. The comma operator effectively simply evaluates each item in sequence and 'returns' the last one.

Quote:Original post by swiftcoder

Quote:Original post by MaulingMonkey
As for how your code is being handled: You're using the comma operator, which means your code behaves something like this:

temp.end();std::string message = temp.end()-Length;

Which is to say, it doesn't compile, since you can't assign a std::string to a single iterator. The comma operator effectively simply evaluates each item in sequence and 'returns' the last one.

As a thought exercise, the comma operator can be overridden to make that syntax work - though I would highly recommend not doing this in real code:
*** Source Snippet Removed ***

If you're going to go and create great evil like that, you might as well go all out with implicit conversions: