c++：栈的应用之表达式求值

Stack.hpp

1. #pragma once
2. template <class T>
3. class Stack{
4. private:
5. T* _array;
6. size_t _capacity;
7. int _topindex;
8. public:
9. Stack()          //构造函数
10. :_array(0)
11. , _capacity(0)
12. , _topindex(-1)
13. {}
14. void Push(const T& x){                         //入栈操作
15. if (_topindex + 1 == _capacity){
16. _capacity = 2 * _capacity + 3;
17. T* tmp = new T[_capacity];
18. if (tmp == NULL){
19. cout << "failed new" << endl;
20. exit(-1);
21. }
22. memcpy(tmp, _array, sizeof(T) * (_topindex + 1));
23. delete _array;
24. _array = tmp;
25. }
26. _array[++_topindex] = x;
27. }
28. void Pop(){                                      //出栈
29. if (_topindex > -1){
30. //cout << _array[_topindex] << endl;
31. _topindex--;
32. }
33. }
34. bool empty(){                  //清空
35. return _topindex = -1;
36. }
37. const T& top(){                             //取栈顶元素
38. //cout << _array[_topindex] << endl;
39. return _array[_topindex];
40. }
41. };
42. main.cpp
43. #include<iostream>
44. #include<string>
45. #include"Stack.hpp"
46. using namespace std;
47. enum Type{
48. ADD,
49. SUB,
50. MUL,
51. DIV,
52. OP_NUM,
53. };
54. struct Cell{
55. Type _type;
56. int num;
57. };
58. Cell RPNExp[] = {
59. OP_NUM, 12,
60. OP_NUM, 3,
61. OP_NUM, 4,
62. ADD, 1,
63. MUL, 1,
64. OP_NUM, 6,
65. SUB, 1,
66. OP_NUM, 8,
67. OP_NUM, 2,
68. DIV, 1,
69. ADD, 1,
70. };
71. int main(){
72. Stack<int> s2;
73. int i = 0;
74. for (int i = 0; i < sizeof(RPNExp) / sizeof(RPNExp[0]);i++){  //取出元素
75. int left, right;             //两个操作数
76. switch (RPNExp[i]._type){        //将栈中数据出栈
77. case ADD:                         //如果是加号，取两个操作数
78. left = s2.top();
79. s2.Pop();
80. right = s2.top();
81. s2.Pop();
82. s2.Push(left + right);         //新的值入栈
83. break;
84. case SUB:                       //如果是减号，取两个操作数
85. left = s2.top();
86. s2.Pop();
87. right = s2.top();
88. s2.Pop();
89. s2.Push(right - left);
90. break;
91. case MUL:                         //如果是乘号，取两个操作数
92. left = s2.top();
93. s2.Pop();
94. right = s2.top();
95. s2.Pop();
96. s2.Push(left * right);
97. break;
98. case DIV:                        //如果是除号，取两个操作数
99. left = s2.top();
100. s2.Pop();
101. right = s2.top();
102. s2.Pop();
103. s2.Push(right / left);
104. break;
105. case OP_NUM:                     //如果数字，入栈
106. s2.Push(RPNExp[i].num);
107. break;
108. default:
109. return -1;
110. }
111. }
112. cout << s2.top() << endl;
113. return 0;
114. }