c语言逆波兰表达式求值

1.源码实现

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int evalRPN(char **tokens, int tokensLen)
{
    int *number = (int *)malloc(sizeof(int) * tokensLen);
    int res = atoi(tokens[0]);
    int u;
    int i = 0, j = 0;

    while(i < tokensLen)
    {
        if(strcmp(tokens[i], "+") != 0 && strcmp(tokens[i], "-") != 0 && strcmp(tokens[i], "*") != 0 && strcmp(tokens[i], "/") != 0)
        {
            number[j] = atoi(tokens[i]);
            j++;
        }
        else
        {
            if(tokens[i][0] == '+')
            {
                u = number[j-1];
                number[j-2] += u;
                j--;
            }
            else if(tokens[i][0] == '-')
            {
                u = number[j-1];
                number[j-2] -= u;
                j--;
            }
            else if(tokens[i][0] == '*')
            {
                u = number[j-1];
                number[j-2] *= u;
                j--;
            }
            else if(tokens[i][0] == '/')
            {
                u = number[j-1];
                number[j-2] /= u;
                j--;
            }
        }

        i++;
    }

    res = number[0];

    free(number);

    number = NULL;

    return res;
}

int main()
{
    char a[10][20] = {"2", "0", "+"};
    char *tokens[10];
    int i;
    int res;

    for(i=0; i<10; i++)
    {
        tokens[i] = a[i];
    }

    res = evalRPN(tokens, 3);

    printf("%d\n", res);

    return 0;
}

2.编译源码

$ gcc -o test test.c -std=c89

3.运行及其结果

$ ./test
2