Question

我试图从命令行中获取两个参数，一个char和一个1和0的int字符串以及x长度的0。我想检查它是否只有5 1和0，如果是的话我会将它们放入数组中。如果长于5，则为10，如10011 10110;在这种情况下，我想将输入解析为两个不同的数组。第一个数字转到A，第二个数字转到B，依此类推，直到我有两个数组，例如a = 10101，b = 01110。

我怎样才能做到这一点？我已经尝试了一个put（c）直到EOF while循环，但似乎无法从中得到任何东西。下面是我的整个程序，最终将数组放入LLL。

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

struct nodeA {
  int dork; 
  struct nodeA * next;
};

typedef struct nodeA d1;

struct nodeB {
  int dork;
  struct nodeB * next;
};

typedef struct nodeB d2;

struct mathOp {
  char operation;
  struct mathOp * next;
};

typedef struct mathOp dorkOp;

int main(int argc, char * argv[]) {
  char a;
  int b;    

  d1 * curr = NULL, * head = NULL;
  dorkOp * curr3 = NULL, * head3 = NULL;
  int i = 0;

  int aA[8] = {0};
  int aB[8] = {0};
  int tempA, tempB;

  if (argc != 3) {
    printf("Program takes <char> <int>\n");
    exit(-1);
  }

  a = atoi (argv[1]);
  b = atoi (argv[2]);

  printf("Test call a = %c, b = %d \n", a, b);

  tempA = b;
  tempB = b;

  for (i = 0; i < 1; i++) {
    curr3 = (dorkOp *)malloc(sizeof(dorkOp));
    curr3->operation = a;
    curr3->next = head3;
    head = curr;
  }

  while (curr3) {
    printf("%c\n", curr3->operation);
    curr3 = curr3->next;
  }

  for (i = 0; i < 1; i++) {
    curr = (d1 *)malloc(sizeof(d1));
    curr->dork = tempA;
    curr->next = head;
    head = curr;
  }

  while (curr) {
    printf("%d\n", curr->dork);
    curr = curr->next;
  }

  return 0;
}

Answer 1

我没有看到要回答的问题，所以这里是OP代码，带有我的评论，前缀为＆＃39; <--＆＃39;并按行缩进。

良好的缩进有助于提高可读性，所以我缩进了代码（并修改了一些结构和变量定义与良好的编码实践保持一致）

以下是编译结果：

50:16: warning: variable 'tempB' set but not used
48:9: warning: unused variable 'aB'
47:9: warning: unused variable 'aA'

我没有在以下代码中修复任何这些项目。顺便说一下，为什么邮政编码不能编译？

顺便说一下：由OP提供的这段代码没有执行OP评论中列出的任何行动。

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

struct nodeA
{
    int dork;
    struct nodeA * next;
};

typedef struct nodeA d1;

struct nodeB
{
    int dork;
    struct nodeB * next;
};

typedef struct nodeB d2;

struct mathOp
{
    char operation;
    struct mathOp * next;
};

typedef struct mathOp dorkOp;


int main(int argc, char * argv[])
{
    char a;
    int b;
    d1 * curr = NULL;
    d1 * head = NULL;

    //d2 * curr2 = NULL;
    //d2 * head2 = NULL;

    dorkOp * curr3 = NULL;
    dorkOp * head3 = NULL;

    int i = 0;
    //int array[10] = {0};

    int aA[8] = {0};
    int aB[8] = {0};

    int tempA;
    int tempB;


    if(argc != 3)
    {
        printf("Program takes <char> <int>\n");
        exit(-1);
    }

    a = atoi (argv[1]);  
    // <-- argv[1] is expected to be a char, 
    //     and 'a' is defined as a char
    //     so the result of atoi() will be 0
    //     and 'a = 0' probably will not yield anything useful 

    b = atoi (argv[2]); 
    // <-- argv[2] is a null terminated string of 1s and 0s, 
    //     which would be easy to parse, so why the conversion? 

    printf("Test call a = %c, b = %d \n", a, b); 
    // <-- 'a' is an unprintable char, so the output is doubtful

    tempA = b;  // <-- shouldn't this be: tempA = a;
    tempB = b;

    for(i = 0; i < 1; i++)  
    // <-- waste of a 'for' because only room for one dorkOp 
    //     and the 'for' will only execute once
    //     suggest removing the 'for' but not the enclosed code block
    {
        curr3 = (dorkOp *)malloc(sizeof(dorkOp));
        // <-- in C, casting the returned value from malloc is 
        //     bad programming practice, for several reasons
        // <-- the returned value from malloc() needs to be checked
        //     to assure successful operation
        //     if not successful, 
        //     then, following lines are dereferencing an offset from 0
        //           which will result in a seg fault event
        curr3->operation = a;
        // <-- 'a' contains 0x00, (see above comment about setting 'a')
        //     so curr3->operation now contains 0x00
        curr3->next = head3; 
        // <-- head3 contains NULL
        //     so this line places NULL in curr3->next
        head = curr;  
        // <-- curr AND head are already both ptrs containing NULL
        //     so this line has no effect
    }

    while(curr3)
    {
        printf("%c\n", curr3->operation);
        // <-- curr3->operation was (see above) set to 0x00
        //     which is a non-printable char
        //     so the displayed output is in doubt
        curr3 = curr3->next;
        // <-- curr3, on the first pass through this loop is set to NULL
    }

    for(i = 0; i < 1; i++)
    // <-- waste of a 'for' because only room for one d1 
    //     and the 'for' will only execute once
    //     suggest removing the 'for' but not the enclosed code block
    {
        curr = (d1 *)malloc(sizeof(d1));
        // <-- in C, casting the returned value from malloc is 
        //     bad programming practice, for several reasons
        // <-- the code needs to check the returned value from malloc
        //     to assure a successful operation
        //     if not successful
        //     then, the following code will be dereferencing from address 0
        //           which will result in a seg fault event
        curr->dork = tempA;
        curr->next = head;
        // <-- head contains NULL so now curr-> contains NULL
        head = curr; 
    }

    // <-- following loop will only execute once
    //     so why make it a loop?
    while(curr)
    {
        printf("%d\n", curr->dork);
        curr = curr->next;
    }

    return 0;
}  // end function: main

如何将命令行参数中的数字解析为两个数组

1 个答案: