![]() ![]() So you ask yourself, “O.K., what’s causing this traffic tie-up?” Your first answer will probably focus on the forces responsible for the heavy traffic, such as “decaying road surfaces” and “frequent accidents.” But, aren’t these patterns and trends you could plot on behavior over time graphs? At this point, you may frown and ask, “So what exactly is structure? What distinctions are we making?”īy its very nature, structure is difficult to see. You know that you can gain leverage and find potential lasting solutions by surfacing the underlying structure. Let’s say that, on your way home from work, you run into a traffic jam. There are guidelines, however, that can help you to see structure and seeing structure will provide you with leverage for solving your long-term business challenges. You are not alone many learners encounter these difficulties. Perhaps when you try to analyze an issue from this perspective, the lines between patterns and structure get blurred or you find structure hard to see. But still, you may be finding it difficult to apply this distinction. This means, that you can find an element that compares equivalent to the given argument.You have probably already learned about the importance of distinguishing among events, patterns, and structure. Since C++14, you can use find() to do a "transparent comparison" as explained on std::set::find(). However, as your operatorkey data s(comp) For your case, this means that you have to provide an instance of test. Prior to C++14, find() expects an element as argument, whose type matches the type of elements stored in the set. However, your actual problem is, that you can't search for parts of your struct by using find() prior to C++14. two objects a and b are considered equivalent if neither compares less than the other: !comp(a, b) & !comp(b, a). Note that in this special case, the constructor must not be declared explicit.įirst and foremost, in order to make std::set::find() work with your struct, you don't need to specify operator=, as explained for std::set: Then passing a string to find() would automatically call the constructor and create a temporary test struct containing only the relevant key. Test(const std::string& strKey = "", const std::string& strData = "") In order to do so, add a constructor for automatic conversion, so the final struct would look like this: struct test Now on to your real problem - your are passing a string to the find() function, but it only accepts structs of type test. ![]() I suggest you operator< and operator= to your struct instead of overloading the global operator, I find it much cleaner example: struct test usr/include/c++/4.2.1/bits/stl_set.h:433:7: note: candidate function not viable: no known conversion from 'const char ' to 'const key_type' (aka 'const test') for 1st argument usr/include/c++/4.2.1/bits/stl_set.h:429:7: note: candidate function not viable: no known conversion from 'const char ' to 'const key_type' (aka 'const test') for 1st argument ![]() In file included from /usr/include/c++/4.2.1/set:65: Here are the error messages that I get when I try to compile it: test.cpp:30:7: error: no matching member function for call to 'find' It works fine except for the part where find() is used: #include I want find() to look only at one of the strings in the struct. However, since the set is holding structs, it doesn't work. I want to be able to use the find() functionality of sets. I am using a set to hold structs which contain several strings. ![]()
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