C++ no instance of constructor matches the argument list

Question

I've been coding a dll and it is almost finished, the final issue I have is this. Argument list consists of: (const char [23], const char [23], const char [21], const char [24], const char [24], const char [24], const char [24], const char [17], const char [42]) The above error displayed at this place in my code...

   std::vector<char*> monitors = { //messages to look out for.  Might as well grab everything fairfight related
    "XiteNetworkGet1Message",
    "XiteNetworkGet2Message",
    "XiteNetworkGoMessage",
    "XiteNetworkInfo1Message",
    "XiteNetworkInfo2Message",
    "XiteNetworkInfo3Message",
    "XiteNetworkInfo4Message",
    "XiteNetworkPush1",
    "ScreenshotNetworkRequestScreenshotMessage"
};

Those char pointers are [const](https://stackoverflow.com/questions/14570993/why-are-string-literals-const). So you need to use `vector`. — Paul Rooney, Oct 07 '20 at 00:46

score 1 · Accepted Answer · answered Oct 07 '20 at 00:46

1

String literals are not modifiable, and therefore cannot be stored as char*. You can use const char* instead:

std::vector<const char*> monitors = { ... };

Or, as pointed out in the comments section, you could consider std::vector<std::string>. That would be preferable if there is no special need to store these literals as pointers. Usually there is no such need.

answered Oct 07 '20 at 00:46

paddy

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For storing a bunch of literals, `std::vector` is the way to go. No need for any dynamic allocation here. – Asteroids With Wings Oct 07 '20 at 01:12

C++ no instance of constructor matches the argument list

1 Answers1