Calculator which can take words as input
I want to write a calculator which can take words as input. for e.g. "two plus five multiply with 7" should give 37 as output. I won't lie, this is a homework so before doing this, I thought if I can be pointed to something which might be useful for these kinds of things and I am not aware of.
Also, approach n how to do this would be ok too , I guess. It has to be written in C++. No other language would be accepted.
Thanks.
[Edit] -- Thanks for the answers. This is an introductory开发者_JS百科 course. So keeping things as simple as possible would be appreciated. I should have mentioned this earlier.
[Edit 2] -- Reached a stage where I can give input in numbers and get correct output with precedence and all. Just want to see how to convert word to number now. Thanks to everybody who is trying to help. This is why SO rocks.
As long as the acceptable input is strict enough, writing a recursive descent parser should be easy. The grammar for this shouldn't differ much from a grammar for a simple calculator that only accepts digits and symbols.
With an std::istream
class and the extraction operator (>>
) you can easily break the input into separate words.
Here's a lecture that shows how to do this for a regular calculator with digits and symbols.
The main difference from that in your case is parsing the numbers. Instead of the symbols '0-9', your input will have words like "zero", "one", "nine", "hundreds", "thousands", etc. To parse such a stream of words you can do something like this:
break the words into groups, nested by the "multipliers", like "thousands", "hundreds", "billions", etc; these groups can nest. Tens ("ten", "twenty", "thirty"), teens ("eleven", "twelve", "thirteen", ...), and units ("one", "two", "three", ...) don't nest anything. You turn something like "one hundred and three thousands two hundred and ninety seven" into something like this:
+--------+---------+-----+ / | | \ thousand hundred ninety seven / \ | hundred three two | one
for each group, recursively sum all its components, and then multiply it by its "multiplier":
103297 +--------+------+----+ / | | \ (* 1000) + (* 100) + 90 + 7 / \ | (* 100) + 3 2 | 1
This was some of my favorite stuff in school.
The right way to do this is to implement it as a parser -- write a grammar, tokenize your input, and parse away. Then you can evaluate it recursively.
If none of that sounds familiar, though (i.e. this is an introductory class) then you can hack it together in a much less robust way -- just do it how you'd do it naturally. Convert the sentence to numbers and operations, and then just do the operations.
NOTE: I'm assuming this is an introductory level course, and not a course in compiler theory. If you're being taught about specific things relevant to this (e.g. algorithms other than what I mention), you will almost certainly be expected apply those concepts.
First, you'll have to understand the individual words. For this purpose, you can likely just hack it together - read one word at a time and try to understand that. Gradually build up a function which can read the set of numbers you need to be able to work with.
If the input is simple enough (only expressions of the basic form you provide, no parentheses or anything), you can simply alternate between reading a number and an operator until the input is fully read (but if this is supposed to be robust, you need to stop and display an error if the last thing you read is an operator), so you can write separate methods for numbers and operators.
To understand how the expression needs to be calculated, use the shunting yard algorithm to parse the expression with proper operator precedence. You can likely combine the initial parsing of the words with this by simply using that to supply the tokens to the algorithm.
To actually calculate the result, you'll need to evaluate the parsed expression. To do that, you can simply use a stack for the output: when you would normally push an operator, you instead pop 2 values, apply the operator to them, and push the result. After the shunting yard algorithm is complete, there will be one value on the stack, containing the final result.
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