I have the following simple expression parser:
expr(+(T,E))-->term(T),"+",expr(E).
expr(T)-->term(T).
term(*(F,T))-->factor(F),"*",term(T).
term(F)-->factor(F).
factor(N)-->nat(N).
factor(E)-->"(",expr(E),")".
nat(0)-->"0".
nat(1)-->"1".
nat(2)-->"2".
nat(3)-->"3".
nat(4)-->"4".
nat(5)-->"5".
nat(6)-->"6".
nat(7)-->"7".
nat(8)-->"8".
nat(9)-->"9".
However this only supports 1-digit numbers. How can I parse numbers with multiple digits in this case?
Use accumulator variables, and pass those in recursive calls. In the following, A and A1 are the accumulator.
digit(0) --> "0".
digit(1) --> "1".
% ...
digit(9) --> "9".
nat(N) --> digit(D), nat(D,N).
nat(N,N) --> [].
nat(A,N) --> digit(D), { A1 is A*10 + D }, nat(A1,N).
Note that the first nat clause initializes the accumulator by consuming a digit, because you don't want to match the empty string.
nat(0).
nat(N):-nat(N-1).
But you use a syntax that I don't know (see my comment above).
Can you provide a sample input?
I think this might work:
nat(N)-->number(N).
If that fails try:
nat(N)-->number(N),!.
The ! is a cut it stops the unification. You can read about it in books/tutorials.
来源：https://stackoverflow.com/questions/3279822/parsing-numbers-with-multiple-digits-in-prolog