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c++autoreturn-type-deduction
Moomin
Moomin

Reputation: 1926

C++ function with variable return type using 'auto'

I'm trying to write a function that returns various types depending on the if-statement.

auto parseParameterValue(QString aParameterValueString, int aParameterType)
{
    if(aParameterType == 0)
    {
        int result = aParameterValueString.toInt();
        return result;
    }
    else if(aParameterType == 1)
    {
        double result = aParameterValueString.toDouble();
        return result; // <------- compilation error
    }
    else
    {
        return aParameterValueString;
    }
}

Unfortunately what I get is:

  1. Warning: 'parseParameterValue' function uses 'auto' type specifier without trailing return type
  2. Error at second return: inconsistent deduction for 'auto': 'int' and then 'double'

Is there a way to make it work?

Thanks in advance.

Upvotes: 3

Views: 2136

Answers (2)

Walter
Walter

Reputation: 45464

You could return an erasure type, such as boost::any, which can store any type. Your code would then look like this

boost::any parseParameterValue(QString aParameterValueString, int aParameterType)
{
    switch(aParameterType) {
    default: return {aParameterValueString};
    case 0:  return {aParameterValueString.toInt()};
    case 1:  return {aParameterValueString.toDouble()};
    }
}

Upvotes: 4

aschepler
aschepler

Reputation: 72431

No, a function can only have one return type.

Note that processing of a function return type must happen at compile time, but your function uses values that can't be known until run time.

Upvotes: 10

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