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struct Bracket {
    Bracket(char type, int position):
        type(type),
        position(position)
    {}

    bool Matchc(char c) {
        if (type == '[' && c == ']')
            return true;
        if (type == '{' && c == '}')
            return true;
        if (type == '(' && c == ')')
            return true;
        return false;
    }

    char type;
    int position;
};
Adrian Mole
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Jazy Matricx
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    It initializes member variables with arguments. What exactly you don't understand? – Daniel Langr Aug 16 '20 at 14:08
  • And it can be done with both curly brackets as parenthesis: `Bracket(char type, int position): type{type}, position{position} {}`. –  Aug 16 '20 at 14:09
  • Possibly relevant question: [Initializing member variables using the same name for constructor arguments as for the member variables allowed by the C++ standard?](https://stackoverflow.com/q/6185020/580083). – Daniel Langr Aug 16 '20 at 14:10
  • You can think about it as `Bracket(char in_type, int in_position) : type(in_type), position(in_position) {}`. It initialises the member variables `type` and `position` of `Bracket` – Ody Aug 16 '20 at 14:14
  • Are you confused that this is a `struct` instead of a `class`? I ask because the member initialization list should be covered when classes are introduced in any `c++` text book. What may not be covered is that a `struct` is the same as a `class` but instead of it defaults to public instead of private. Related: [https://stackoverflow.com/a/999810/487892](https://stackoverflow.com/a/999810/487892) – drescherjm Aug 16 '20 at 14:14

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