Interesting approach!
I've personally done something similiar in a project a few weeks ago. First, I'm taking the input in infix notation (1 + 2) and convert it into RPN (reverse polish notation, 1 2 +) using Dijkstra's Shunting-yard algorithm.
1 + 2
1 2 +
Having the expression in RPN format, you start solving it by pushing values and operators on a Stack and pop()-ing through them, until no more data to process is available.
pop()
You can see my code over here on Github inside the project I used it for: github.com/RicoBrase/ChatCalculato...
Asscociated unit test: github.com/RicoBrase/ChatCalculato...
There are some code leftovers from experimenting with Optionals, just ignore that. 😉
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Interesting approach!
I've personally done something similiar in a project a few weeks ago.
First, I'm taking the input in infix notation (
1 + 2) and convert it into RPN (reverse polish notation,1 2 +) using Dijkstra's Shunting-yard algorithm.Having the expression in RPN format, you start solving it by pushing values and operators on a Stack and
pop()-ing through them, until no more data to process is available.You can see my code over here on Github inside the project I used it for:
github.com/RicoBrase/ChatCalculato...
Asscociated unit test:
github.com/RicoBrase/ChatCalculato...
There are some code leftovers from experimenting with Optionals, just ignore that. 😉