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一.函数实现目标

写一个Matlab函数 function ret = myeval(str)计算表达式str的值。

str是一个算术表达式，即两个正整数的加法，减法，乘法和除法。

例如，myeval(’ 12+23 ‘)将执行返回35,myeval(’ 23-12 ‘)将返回11,myeval(’ 12*11 ‘)将返回132，myeval(’ 12/4 ‘)将返回3。

二.额外要求

(1) 其中可以多一个空格表达式，应该忽略它。

例如，“12 + 23”等于“12 + 23”。

(2)不能使用Matlab函数eval或num2str等现有的函数。

三.整体思路

核心思想：依靠ASCII码解决



step1

：将输入的字符串转化成ASCII码


step2

：检测有无空格，并将除了空格之外的字符的ASCII码保存到 ls[]


step3

：迭代 ls[]，将第一个数的各个位保存到num1[]，当检测到操作符时，将其保存到operator[]， 并且跳出循环。


step4

: 将第二个数的(ls[]中除了第一个数和操作符)各个位保存到num2[]


step5

: 根据num1和num2计算这两个数的真实值Number1， Number2



根据ASCII码表，我们可以知道字符0对应的ASCII码是48，所以将字符的ASCII码减去48即可得到各个位的真实值，然后再根据在哪一位，计算出两个数字的真实值。


step6

: 根据操作符operator[]判断是哪种运算并计算出结果rat

四.完整代码

function ret = myeval(str)
% MYEVAL: calculate the addition, subtraction, multiplication and division of two Numbers entered as a string 
% The core idea is to use ASCII code to distinguish between two numeric and operator types 

% Converts the input string to ASCII code
str = abs(char(str));
ls = [];
% The number of number of the ASCII code of the string except space and set it to 1 initially
lsEND = 1;
for i = str
   % All but Spaces (32 for ASCII code) are added to ls
   if i ~= 32
       ls(lsEND)= i ;
       lsEND = lsEND + 1;
   end
end

num1 = [];
% The number of number of the first number and set it to 1 initially
num1END = 1;
operator = [];
for i = ls
   % Store each digit of the first number in num1
   if i < 48 || i > 57
       % Save the ASCII code for the operator and jump out of the for-loop
       operator(1) = i;
       break
   end
   num1(num1END) = i;
   num1END = num1END + 1;
  
end

% Calculate the true value of the first number
Number1 = 0;
for m = 1:(num1END - 1)
    % The ASCII - 48 of the numeric character gets the true value of the number, and then multiplies the corresponding number of digits
    Number1 = Number1 + (num1(m) - 48) * 10^(num1END - 1 - m);
end

num2 = [];
% The number of number of the second number and set it to 1 initially
num2END = 1;
for j = ls(num1END + 1) : ls(end)
   % Store each digit of the second number in num2
   num2(num2END) = j;
   num2END = num2END + 1;
end

% Calculate the true value of the second number
Number2 = 0;
for n = 1:(num2END - 1)
    % The ASCII - 48 of the numeric character gets the true value of the number, and then multiplies the corresponding number of digits
    Number2 = Number2 + (num2(n) - 48) * 10^(num2END - 1 - n);
end

% According to the ASCII code of the operator, what operation is to be performed and operate
% '+': 43; '-': 45; '*': 42; '/': 47;
switch operator(1)
    case 43
        ret = Number1 + Number2;
    case 45
        ret = Number1 - Number2;
    case 42
        ret = Number1 * Number2;
    case 47
        ret = Number1 / Number2;
end
end