C++ variable arguments using initializer lists

Question

Assume the following code, which is a tiny sprintf substitute. (The _itoa and the likes was just used to keep the code short.)

#include <cstdlib>
#include <string>
class Arg {
public:
    Arg(const std::string& s) :m_str(s) {}
    Arg(const char* s) : m_str(s) {}
    Arg(int digi, double number)  {char buf[128]; m_str = _gcvt(number, digi, buf);}
    operator const std::string& ()const { return m_str; }
private:
    std::string m_str;
};

class Format {
public:
    Format(/*const char* format, */std::initializer_list<Arg> args); // see below
    const std::string& str()const { return m_str; }
private:
    std::string m_str;
};

Format::Format(/*const char* format, */std::initializer_list<Arg> args) {
    auto arg = args.begin();
    auto format = std::string(*arg++);
    for(const char* c = format.c_str(); *c!='\0'; ++c) {
        if(*c=='%') { m_str+=*arg++; }
        else { m_str+=*c; }
    }
}


int main() {

    std::string test1 = Format{"test Double:% String:%", {5, 456.78}, "foo"}.str();

    // I want to make this work. See the braces.
    std::string test2 = Format("test Double:% String:%", {5, 456.78}, "foo").str();

    return 0;
}

You see, I want to pass arguments, limited to type "Arg", but use a constructor that uses e.g. varadic templates instead of the initializer_list<> for better readability.

I tried:

    template<typename... T>
    Format(T&& ... args) : Format(std::forward<Args>(args)...) {}

But i get:

error C2440: '<function-style-cast>': cannot convert from 'initializer list' to 'Format'
note: No constructor could take the source type, or constructor overload resolution was ambiguous

Jarod42 · Accepted Answer · 2020-04-29 11:57:52Z

1

std::initializer_list requires {} not ().

{5, 456.78} has no type and couldn't be deduced for template.

The way to keep your syntax in the old overload way:

Format(Arg arg0) : Format(std::initializer_list{arg0});
Format(Arg arg0, Arg arg1) : Format({arg0, arg1});
Format(Arg arg0, Arg arg1, Arg arg2) : Format({arg0, arg1, arg2});
// ... Up to some limit

edited Apr 29, 2020 at 11:57

answered Apr 29, 2020 at 9:49

Jarod42

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1 Comment

KungPhoo Over a year ago

While this seems like a hack, it does exaclty what I want it to do. Thank you.

songyuanyao · Accepted Answer · 2020-04-29 09:43:27Z

1

Firstly, you should use braces in member initializer list, to forward to the constructor taking std::initializer_list.

template<typename... T>
Format(T&& ... args) : Format{std::forward<T>(args)...} {}
//                           ^                        ^

Secondly, given Format("test Double:% String:%", {5, 456.78}, "foo"), unfortunately braced-init-list like {5, 456.78} can't be deduced in template type deduction, it has no type. You can specify the type explicitly like

std::string test2 = Format("test Double:% String:%", Arg(5, 456.78), "foo").str();
//                                                      ^         ^

answered Apr 29, 2020 at 9:43

songyuanyao

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1 Comment

KungPhoo Over a year ago

Thank you for the excellent explaination of "why does this fail" (template deduction).

ForEveR · Accepted Answer · 2020-04-29 09:42:47Z

0

You can do like this:

template<typename... T>
Format(T&& ... args) : Format({std::forward<T>(args)...}) {}}

But in call you should manually point that second argument is Arg, like this:

std::string test2 = Format("test Double:% String:%", Arg{5, 456.78}, "foo").str();

example

answered Apr 29, 2020 at 9:42

ForEveR

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