Write a function that calculates the result of an arithmetic

Write a function that calculates the result of an arithmetic expression involving scalars, where the arithmetic expression is given as a character string (e.g., \'2*30+5*4\'). The operators allowed in the arithmetic expression are +, - */, and, which represent addition, subtraction, multiplication, division, and exponentiation, respectively However, the order of operations has been changed! Addition and subtraction now have precedence over multiplication and division, which now have precedence over exponentiation Addition and subtraction still have equal precedence Similarly, multiplication and division still have equal precedence. Operators of equal precedence should be evaluated from left to right. Write a function with the following header: function [result] = my_calculator_inverse_precedence(expression) where: expression is a row vector of class char that represents an arithmetic expression as described above. result is a scalar of class double that represents the value of the arithmetic expression described by expression You can assume that: expression is not empty. expression is a valid arithmetic expression. expression contains only characters among: 0123456789.+-*/\". In particular, expression does not contain spaces nor parentheses. The _rst character in expression is one of the 10 digits. There are no two operators in a row (e g. +-) in expression You may not use Matlab\'s built-in functions split, strsplit, find, strfind, eval, and feval in this question.

Solution

Matlab function my_calculator_inverse_precedence.m for the problem

function [result] = my_calculator_inverse_precedence(expression)
   stk_front = 0; % variable indicating the front of stack
   st = \'\'; % empty stack
   op_stk = []; % operator position monitor stack
   op_pos = 0; % operator position stack front
   N = length(expression);% length of the expression
   for k = 1:N % loop to monitor each element in expression
       % checking expression(k) is a operator or not
       if(expression(k) == \'+\' || ...
          expression(k) == \'-\' || ...
          expression(k) == \'*\' || ...
          expression(k) == \'/\' || ...
          expression(k) == \'^\' )
             if(op_pos == 0) % no operator loaded in stack
                 stk_front = stk_front+1;
                 st(stk_front) = expression(k);
                 op_pos = op_pos +1;
                 op_stk(op_pos) = stk_front;
             else % stack have some operator
                 while( pre(st(op_stk(op_pos))) >= pre(expression(k))) % checking the precedence of operators
                     if(op_pos == 1) % one operator
                         C = num2str(str2num(st(1:stk_front))); % evaluating part of the expression
                         st(1:length(C)) = C; % updating stack with evaluated expression
                         stk_front = length(C);
                         op_pos = 0;
                         break;
                     else % Many operator
                         C = num2str(str2num(st(op_stk(op_pos-1)+1:stk_front)));% evaluating part of the expression
                         st(op_stk(op_pos-1)+1:op_stk(op_pos-1)+length(C)) = C; % updating stack with evaluated expression
                         stk_front = op_stk(op_pos-1)+length(C);
                         op_pos = op_pos-1;
                     end
                 end% adding new operotor in the stack
                    stk_front = stk_front+1;
                    st(stk_front) = num2str(expression(k));
                    op_pos = op_pos +1;
                    op_stk(op_pos) = stk_front;
             end
       % checking expression(k) is operand
       else
       % if yes add it in the stack
           stk_front = stk_front+1;
           st(stk_front) = num2str(expression(k));
       end
   end
   % evaluating the expression
   result = str2num(st(1:stk_front));
end
function val= pre(oper)
switch (oper)
         case \'+\'
             val = 2;
         case \'-\'
            val = 2;
         case \'*\'
             val = 1;
         case \'/\'
             val = 1;
         case \'^\'
            val = 0;
end
end

testing the function my_calculator_inverse_precedence.m

>> x = my_calculator_inverse_precedence(\'2*30+5^4\')

x =

    24010000

>> (2*(30+5))^4

ans =

    24010000

>> class(x)

ans =

double

>> x = my_calculator_inverse_precedence(\'4-3.14\')

x =

    0.8600

>> x = my_calculator_inverse_precedence(\'4-2-2\')

x =

     0

>>