Question

我已经尝试过使用atoi然后将它们切换回字符串以进行推送，我正在尝试为类创建一个rpn计算器，因此推送，弹出，搜索和堆栈结构是如何需要的但是我不能让它添加整数值。

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

struct stack
{
  const char * data;
  struct stack *bottom;
};

struct stack * push(struct stack *stk,const char * x)
{
    struct stack *nstk = (struct stack *)malloc(sizeof(struct stack));

    nstk->data = x;
    nstk->bottom = stk;

    return nstk;
}


const char * peek(struct stack *stk)
{
    if(stk -> data)
      return stk->data;
    else
      return("Stack is empty");
}

struct stack *pop(struct stack *stk)
{
    struct stack *tmp;

    tmp = stk->bottom;
    stk->bottom = NULL;
    free(stk);

    return tmp;
}

FILE * input_from_args(int argc,const char *argv[])
{
    if(strcmp(argv[1],"-e") != 0 && strcmp(argv[1],"-c") != 0 && strcmp(argv[1],"-g") != 0)
    {
      printf("Option %s is not supported \n", argv[1]);
      exit(0);
    }

     else
    {
        return stdin;
    }
}
void evaluate(struct stack * equation)
{
     int op;
     int op2;
     int ans;

     if(strcmp("A",equation->data) == 0)
      {
      op = (int)pop(equation)-> data;
      op2 = (int)pop(equation)-> data;
      ans = op + op2;
      printf("%i",ans);
      }
}

void convert(struct stack  * equation)
{
}

void other (struct stack * equation)
{

}


int main(int argc,const char *argv[])
{
  FILE *src = input_from_args(argc, argv);

    if (src == NULL)
    {
        printf("%s", "Invalid Source");
        exit(EXIT_FAILURE);
    }

    struct stack * equation = NULL;

    int i;
    for(i=2; i <= argc; i++)
    {
      equation = push(equation,argv[i]);
    }

    if(strcmp(argv[1],"-e") == 0)
    {
      evaluate(equation);
    }
    else if(strcmp(argv[1],"-c") == 0)
    {
      convert(equation);
    }
    else if(strcmp(argv[1],"-g") == 0)
    {
      other(equation);
    }

    return EXIT_SUCCESS;
}

这就是我到目前为止所有的一切，如果你发现任何其他问题都很好但我真正想知道的是如何用这个数据结构来评估一个后缀方程式，输入的例子是-e 2 2 A 5 X.

Answer 1

我希望会有所帮助。

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

typedef int DataType;

DataType cnv(const char *s){
    return atoi(s);
}

#define PRN_DATA "%d"

typedef struct stack {
    DataType data;
    struct stack *bottom;
} Stack;

void push(Stack **stk, DataType x){
    struct stack *nstk = malloc(sizeof(Stack));

    nstk->data = x;
    nstk->bottom = *stk;

    *stk = nstk;
}

int empty(Stack *stk){
    return stk == NULL;
}

DataType pop(Stack **stk){
    struct stack *tmp = *stk;
    if(empty(*stk)){
        printf("empty stack\n");
        exit(EXIT_FAILURE);
    }
    DataType ret = tmp->data;

    *stk = (*stk)->bottom;
    free(tmp);

    return ret;
}

char input_from_args(int *argc, const char ***argv){
    if( *argc < 2 || (*argv)[1][0] != '-' || (*argv)[1][1] == '\0'){
        printf("Option is not specified\n");
        exit(EXIT_FAILURE);
    }
    char op = (*argv)[1][1];
    if((*argv)[1][2] != '\0' || op != 'e'  && op != 'c' && op != 'g'){
        printf("Option %s is not supported \n", (*argv)[1]);
        exit(EXIT_FAILURE);
    }
    *argv = &(*argv)[2];
    *argc -= 2;
    return op;
}

void evaluate(int argc, const char **argv){
    struct stack *s = NULL;
    int i;
    DataType v1, v2;
    for(i = 0; i < argc; ++i){
        switch(*argv[i]){
        case 'A':
            v2 = pop(&s);
            v1 = pop(&s);
            push(&s, v1 + v2);
            break;
        case 'S':
            v2 = pop(&s);
            v1 = pop(&s);
            push(&s, v1 - v2);
            break;
        case 'X':
            v2 = pop(&s);
            v1 = pop(&s);
            push(&s, v1 * v2);
            break;
        case 'D':
            v2 = pop(&s);
            v1 = pop(&s);
            push(&s, v1 / v2);
            break;
        default:
            push(&s, cnv(argv[i]));
        }
    }
    printf(PRN_DATA "\n", pop(&s));
    if(!empty(s)){
        printf("data remains in the stack\n");
        exit(EXIT_FAILURE);
    }
}

void convert(int argc, const char **argv){
}

void other (int argc, const char **argv){
}

int main(int argc, const char **argv){
    switch(input_from_args(&argc, &argv)){
    case 'e':
        evaluate(argc, argv);
        break;
    case 'g':
        other(argc, argv);
        break;
    case 'c':
        convert(argc, argv);
        break;
    }

    return EXIT_SUCCESS;
}

我如何添加值，因为它们是字符串？

1 个答案: