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使用 typeinfo 获取类名称，实现简单字符串模版参数，这种用法有什么弊端吗？

 2 years ago
source link: https://www.v2ex.com/t/817538
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V2EX  ›  C++

使用 typeinfo 获取类名称，实现简单字符串模版参数，这种用法有什么弊端吗？

  sl0000 · 1 天前 · 612 次点击 

#include <cxxabi.h>

template <typename T>
static inline const std::string classname() {
    int status;
    std::string actual_class_name = abi::__cxa_demangle(typeid(T).name(), 0, 0, &status);
    if (status != 0) {
        throw "Class " + std::string(typeid(T).name()) + " parse error";
    } else {
        return actual_class_name;
    }
}


template <typename T>
class kind {
    static inline const std::string n = classname<T>();
};

class dog { };
name<dog> d;

class cat { };
name<cat> c;

第 1 条附言  ·  22 小时 5 分钟前

@GeruzoniAnsasu c++2a 可以使用字符串模版

template<size_t N>
struct StringLiteral {
    
    constexpr StringLiteral(const char (&str)[N]) {
        std::copy_n(str, N, value);
    }
    
    char value[N];
};

template<StringLiteral lit>
void Print() {
    // The size of the string is available as a constant expression.
    constexpr auto size = sizeof(lit.value);
    // and so is the string's content.
    constexpr auto contents = lit.value;
    std::cout << "Size: " << size << ", Contents: " << contents << std::endl;
}

int main(int argc, const char * argv[]) {
    Print<"literal string">(); // Prints "Size: 15, Contents: literal string"
    return 0;
}

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