Compilers differ in operator precedence

[sblionel]

@FortranFan

The term per the standard is defined operation , specifically a binary defined operation which the standard labels as defined-binary-op .

So even though the form you are employing is x1 intrinsic-operator x2 , it still remains a defined-binary-op .

I don’t agree with this at all.

R1023 defined-binary-op is . letter [ letter ] … .

There is no way that // can be interpreted as a defined-binary-op. These are syntax rules, and // is its own syntax term that appears in:

R1010 level-3-expr is [ level-3-expr concat-op ] level-2-expr
R1011 concat-op is //

That you extended the intrinsic with your own generic procedure does not magically make the intrinsic operator a defined-binary-op. Therefore, its precedence is unchanged.

I talk about operator precedence in Doctor Fortran in “Order! Order!” - Doctor Fortran (stevelionel.com)