I know I can write:

SpokenString[a + b^2]

And then obtain as output:

"a plus b squared"

What about the opposite, something like:

ReadString["a plus b squared"]

And then obtain as output (without access to the internet, e.g., calling Wolfram Alpha):

a + b^2
  • 2
    $\begingroup$ SpokenString? $\endgroup$
    – kglr
    Commented Apr 18, 2018 at 22:59
  • $\begingroup$ Yes, found it right after. Changed my question to ask if there is a way to do the reverse without calling WolframAlpha $\endgroup$
    – Luxspes
    Commented Apr 18, 2018 at 23:01
  • $\begingroup$ SpokenString is a great suggestion to go from code -> prose. If you want to go from prose -> code, I recommend checking out inputs in wolframalpha, like wolframalpha.com/input/…. $\endgroup$
    – enano9314
    Commented Apr 18, 2018 at 23:01
  • $\begingroup$ @Luxspes I unfortunately seemed to have missed your edit by a few minutes. If you don't want to call W|A I am not sure your problem is solvable without MUCH work $\endgroup$
    – enano9314
    Commented Apr 18, 2018 at 23:02
  • 1
    $\begingroup$ if you want to restrict input to a handful of "operators" maybe you could do it. Otherwise to be really general you would pretty much be reinventing alpha. $\endgroup$
    – george2079
    Commented Apr 19, 2018 at 0:00

1 Answer 1



without calling Wolfram Alpha

means you just want an in-product approach as opposed to a web based approach, you could use Ctrl+= or the WolframAlpha function. For example:

WolframAlpha["a plus b squared", "MathematicaParse"]

HoldComplete[a + b^2]

If you really mean that you don't want to use anything based on Wolfram|Alpha, then I think you will need to create your own parser.

  • $\begingroup$ I just want to know if a local function to do it exists $\endgroup$
    – Luxspes
    Commented Apr 19, 2018 at 0:02
  • 1
    $\begingroup$ @Luxspes Meaning you want a function that you can use without access to the internet? If so, I think you're out of luck. $\endgroup$
    – Carl Woll
    Commented Apr 19, 2018 at 0:29
  • $\begingroup$ Is there a way to be sure? $\endgroup$
    – Luxspes
    Commented Apr 19, 2018 at 1:36

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