C++ initialization of reference variable

Question

I spotted a "mistake" in my code today... But I'm not sure if the "mistake" actually changes the resulting compiled code.

Consider initialization of the following reference to a double , x .

double &x{*_data->x_double}; // x_double is a member variable of the struct
                             // _data, it is a pointer to another double

// eg, I have this somewhere else...
struct data
{
    double *x_double;
};

data *_data = new data; // edit: duh, this has to be a pointer...

double another_x = 10.0;
_data->x_double = &another_x; // edit: pointer here too!

However I made the following "mistake"... (notice the extra = sign)

double &x={*_data->x_double};

This code is a minimal example copied from my actual code, in which I don't have references to doubles, but references to large objects such as std::vector 's.

The reason for the reference variables is that they are used in an algorithm, and the variable names are very long, so I create shorted aliases for those variables using references. Hope it makes sense why I did that...

So, I've "corrected" the "mistake" but has my output compiled code actually changed?

Answer 1

They're both list initialization . For your cases:

double &x={*_data->x_double};

Is :

6) initialization of a named variable with a braced-init-list after an equals sign

And

double &x{*_data->x_double};

Is :

1) initialization of a named variable with a braced-init-list (that is, a possibly empty brace-enclosed list of expressions or nested braced-init-lists)

Their effects are in this case exactly the same.