在 C 中标记字符串?

问题描述

我正在为用 C 编写的计算器开发终端解析器.我不知道如何连接运算符之间的所有数字以将它们放入数组中.

I'm working on a terminal parser for a calculator written in C.I cannot figure out how to concatenate all of the numbers that are in between operators to put them into an array.

例如，如果输入(命令行参数)是4+342"，理想情况下应该是 input[] = {"4", "+", "342"}.

For example, if the input (command line argument) was "4+342",it would ideally be input[] = {"4", "+", "342"}.

到目前为止，这是我的代码.我包括 、 和 .

Here's my code so far. I'm including <stdio.h>, <stdlib.h>, and <ctype.h>.

typedef char * string;

int main(int argc, char *argv[])
{
  string inputS = argv[1];
  string input[10];
  string temp;
  printf("%s\n", inputS);
  int i;
  int len = strlen(inputS);
  printf("parsed:\n");
  for(i = 0; i < len; inputS++, i++)
  {
    if(isdigit(*inputS))
    {
      printf("%c",*inputS);
    }
    else
    {
      printf("\n%c\n",*inputS);
    }
  }
  printf("\n");
  return 0;
}

如果用./calc 4+5-546运行，会输出:

4
+
5
-
546

那么将每一行放到它自己的数组槽中的最简单方法是什么?

So what's the easiest way to get each line of this into its own array slot?

推荐答案

试试这个大小...

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

typedef char * string;

int main(int argc, char *argv[])
{
    string inputS = argv[1];
    string input[50];   /* Up to 50 tokens */
    char   buffer[200];
    int    i;
    int    strnum = 0;
    char  *next = buffer;
    char   c;

    if (argc != 2)
    {
        fprintf(stderr, "Usage: %s expression\n", argv[0]);
        return 1;
    }

    printf("input: <<%s>>\n", inputS);
    printf("parsing:\n");

    while ((c = *inputS++) != '\0')
    {
        input[strnum++] = next;
        if (isdigit(c))
        {
            printf("Digit: %c\n", c);
            *next++ = c;
            while (isdigit(*inputS))
            {
                c = *inputS++;
                printf("Digit: %c\n", c);
                *next++ = c;
            }
            *next++ = '\0';
        }
        else
        {
            printf("Non-digit: %c\n", c);
            *next++ = c;
            *next++ = '\0';
        }
    }

    printf("parsed:\n");
    for (i = 0; i < strnum; i++)
    {
        printf("%d: <<%s>>\n", i, input[i]);
    }

    return 0;
}

给定程序被称为 tokenizer 和命令:

Given the program is called tokenizer and the command:

tokenizer '(3+2)*564/((3+4)*2)'

它给了我输出:

input: <<(3+2)*564/((3+4)*2)>>
parsing:
Non-digit: (
Digit: 3
Non-digit: +
Digit: 2
Non-digit: )
Non-digit: *
Digit: 5
Digit: 6
Digit: 4
Non-digit: /
Non-digit: (
Non-digit: (
Digit: 3
Non-digit: +
Digit: 4
Non-digit: )
Non-digit: *
Digit: 2
Non-digit: )
parsed:
0: <<(>>
1: <<3>>
2: <<+>>
3: <<2>>
4: <<)>>
5: <<*>>
6: <<564>>
7: <</>>
8: <<(>>
9: <<(>>
10: <<3>>
11: <<+>>
12: <<4>>
13: <<)>>
14: <<*>>
15: <<2>>
16: <<)>>

这篇关于在 C 中标记字符串?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！