I found this remark: https://www.cs.fsu.edu/~lacher/courses/COT4401/notes/cise_v2_i1/fortran.pdf https://www.cs.fsu.edu/~lacher/courses/COT4401/notes/cise_v2...

I found this remark: https://www.cs.fsu.edu/~lacher/courses/COT4401/notes/cise_v2_i1/fortran.pdf https://www.cs.fsu.edu/~lacher/courses/COT4401/notes/cise_v2... The Fortran I compiler would expand each operator with a sequence of parentheses. In a simplified form of the algorithm, it would • replace + and – with ))+(( and ))-((, respectively; • replace * and / with )*( and )/(, respectively; • add (( at the beginning of each expression and after each left parenthesis in the original expression; and • add )) at the end of the expression and before each right parenthesis in the original expression. This is the first time I hear about this, this technique will work with today compilers vs. most others like precedence climb?