用C语言帮我写一个表达式求值问题的代码

红太狼 2024-03-25 17:54 46阅读 0赞

好的，以下是一个用C语言编写的基本表达式求值问题的代码示例：

\`\`\` \#include <stdio.h> \#include <stdlib.h>

// 定义栈结构体 typedef struct \{ int \*array; // 栈内元素数组指针 int size; // 栈的容量 int top; // 栈顶指针 \} Stack;

// 初始化栈 void initStack(Stack *s, int size) \{ s->array = (int*) malloc(size \* sizeof(int)); s->size = size; s->top = -1; \}

// 判断栈是否为空 int isEmpty(Stack \*s) \{ return s->top == -1; \}

// 判断栈是否已满 int isFull(Stack \*s) \{ return s->top == s->size - 1; \}

// 入栈 void push(Stack \*s, int data) \{ if (isFull(s)) \{ printf("Error: Stack is full.\\n"); return; \} s->top++; s->array\[s->top\] = data; \}

// 出栈 int pop(Stack \*s) \{ if (isEmpty(s)) \{ printf("Error: Stack is empty.\\n"); return -1; \} int data = s->array\[s->top\]; s->top--; return data; \}

// 获取栈顶元素 int peek(Stack \*s) \{ if (isEmpty(s)) \{ printf("Error: Stack is empty.\\n"); return -1; \} return s->array\[s->top\]; \}

// 判断操作符优先级 int precedence(char op) \{ if (op == '+' || op == '-') \{ return 1; \} else if (op == '\*' || op == '/') \{ return 2; \} else \{ return 0; \} \}

// 计算表达式 int evaluate(char *expression) \{ Stack operandStack; // 操作数栈 Stack operatorStack; // 操作符栈 int i = 0; while (expression\[i\] != '\\0') \{ if (expression\[i\] == ' ') \{ // 忽略空格 i++; continue; \} else if (isdigit(expression\[i\])) \{ // 如果是数字，将其入栈 int operand = 0; while (isdigit(expression\[i\])) \{ operand = operand \* 10 + (expression\[i\] - '0'); i++; \} push(&operandStack, operand); \} else \{ // 如果是操作符，将其入栈 while (!isEmpty(&operatorStack) && precedence(peek(&operatorStack)) >= precedence(expression\[i\])) \{ int operand2 = pop(&operandStack); int operand1 = pop(&operandStack); char op = pop(&operatorStack); if (op == '+') \{ push(&operandStack, operand1 + operand2); \} else if (op == '-') \{ push(&operandStack, operand1 - operand2); \} else if (op == '*') \{ push(&operandStack, operand1 \* operand2); \} else if (op == '/') \{ push(&operandStack, operand1 / operand2); \} \} push(&operatorStack, expression\[i\]); i++; \} \} while (!isEmpty(&operatorStack)) \{ int operand2 = pop(&operandStack); int operand1 = pop