How to deduce the type of lambda's return value.?

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As per the documatation(What is a lambda expression in C++11?), The return type of the lambda could be deduced in this code. I could not get the idea how it could be done?

    void func4(std::vector<double>& v) 
    {
      std::transform(v.begin(), v.end(), v.begin(),
                     [](double d) { return d < 0.00001 ? 0 : d; }
                     );
    }

But,it cannot be deduced in this example, why?

 void func4(std::vector<double>& v) {
        std::transform(v.begin(), v.end(), v.begin(),
            [](double d) {
                if (d < 0.0001) {
                    return 0;
                } else {
                    return d;
                }
            });
    }
c++11 c++
1 Answers

Sloppy speaking, the ternary operator has some built-in conversion to a common type. You can read on cppreference how the type of the result is determined. The details are rather involved, so i'll put it in plain English: The result of d < 0.00001 ? 0 : d; is double.

In the no-conditional version one branch returns an int the other a double, hence the return type cannot be deduced.

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