用算符优先法对4+2*3-10/5求值,用c语言写出来
时间: 2024-03-25 07:40:57 浏览: 24
算符优先法的求值步骤如下:
1. 将表达式转换成后缀表达式
2. 用栈存储数字和运算符
3. 从左到右扫描后缀表达式
4. 如果是数字,入栈
5. 如果是运算符,弹出栈顶的两个数字进行计算,并将结果入栈
对于表达式4+2*3-10/5,转换成后缀表达式为4 2 3 * + 10 5 / -
C语言代码如下:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_SIZE 100
typedef struct {
int top;
int data[MAX_SIZE];
} Stack;
void push(Stack *s, int value) {
if (s->top == MAX_SIZE - 1) {
printf("Stack overflow!\n");
exit(1);
}
s->top++;
s->data[s->top] = value;
}
int pop(Stack *s) {
if (s->top == -1) {
printf("Stack underflow!\n");
exit(1);
}
int value = s->data[s->top];
s->top--;
return value;
}
int is_operator(char c) {
return c == '+' || c == '-' || c == '*' || c == '/';
}
int precedence(char c) {
if (c == '+' || c == '-') {
return 1;
} else if (c == '*' || c == '/') {
return 2;
} else {
return 0;
}
}
int evaluate(char *expression) {
Stack stack;
stack.top = -1;
char *token = strtok(expression, " ");
while (token != NULL) {
if (!is_operator(token[0])) {
push(&stack, atoi(token));
} else {
int operand2 = pop(&stack);
int operand1 = pop(&stack);
switch (token[0]) {
case '+':
push(&stack, operand1 + operand2);
break;
case '-':
push(&stack, operand1 - operand2);
break;
case '*':
push(&stack, operand1 * operand2);
break;
case '/':
push(&stack, operand1 / operand2);
break;
default:
printf("Invalid operator: %c\n", token[0]);
exit(1);
}
}
token = strtok(NULL, " ");
}
return pop(&stack);
}
int main() {
char expression[] = "4 2 3 * + 10 5 / -";
int result = evaluate(expression);
printf("Result: %d\n", result);
return 0;
}
```
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