回答編集履歴
1
追記
test
CHANGED
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@@ -48,4 +48,59 @@
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48
48
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}
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49
49
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}
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50
50
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```
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51
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**追記**
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52
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演算子の優先順位と括弧の処理も追加した構文解析です。
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53
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```Java
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54
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import java.util.Scanner;
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51
55
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56
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class Main {
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57
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public static void main(String[] args) {
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58
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System.out.println("数式を入力してください");
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59
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Scanner scanner = new Scanner(System.in);
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60
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String formula = scanner.nextLine();
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61
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Expr e = new Expr(formula);
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62
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int val = e.syn(0);
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63
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if (e.tok != 0) System.out.println("syntax error");
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64
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else System.out.println(val);
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65
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}
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66
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}
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67
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68
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class Expr { // 数式クラス (expression)
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69
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String str; int len, pos, val; char tok;
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70
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71
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Expr(String s) { str = s; len = s.length(); pos = 0; }
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72
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73
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char lex() { // 字句解析 (lexical analyzer)
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74
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while (pos < len)
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75
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if ((tok = str.charAt(pos++))>='0' && tok<='9') {
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76
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for (val = tok - '0'; pos < len &&
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77
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(tok = str.charAt(pos))>='0' && tok<='9'; pos++)
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78
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val = val * 10 + (tok - '0');
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79
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return tok = '0';
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80
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}
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81
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else if (tok != ' ') return tok;
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82
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return tok = 0; // end of string
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83
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}
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84
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85
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final static char[] op = { '+', '-', '*', '/' };
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86
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87
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int syn(int i) { // 構文解析 (syntax analyzer)
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88
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int v;
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89
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if (i < 4)
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90
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for (v = syn(i+2); tok == op[i] || tok == op[i+1]; )
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91
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if (tok == '+') v += syn(i+2);
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92
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else if (tok == '-') v -= syn(i+2);
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93
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else if (tok == '*') v *= syn(i+2);
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94
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else v /= syn(i+2);
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95
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else if (lex() == '0') { v = val; lex(); }
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96
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else if (tok == '(') {
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97
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v = syn(0);
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98
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if (tok == ')') lex();
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99
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else tok = 1;
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100
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}
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101
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else v = tok = 1;
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102
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return v;
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103
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}
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104
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}
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105
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```
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106
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