为什么我的数组前面的元素在变化（C中的字符串数组）

我正在构建一个shell历史函数，以接收字符串，并在程序运行时将其添加到字符串数组中

我的问题是，每当我用新行（字符串）更新数组时，缓存中的前一个元素就会被我的CWD（当前工作目录）填充，但我需要保留我设置的前一个字符串

这是我的主循环，它获取字符串并尝试使用缓存函数存储历史记录：

//prints out the cwd; then loops to take in the line, split it up into arguments, and attempt to execute it
//while lsh_execute returns 0, then frees up the allocated space
void lsh_loop(void)
{
  char *line;               //pointer to a char (the beg. of an array of chars)
  char *cache[10] = {NULL};     //history array
  char **args;              //pointer to a pointer of a char...
  int status, counter = 0, i, j;

  do {
    printf("%s>", getcwd(0,0));     //print cwd
    line = lsh_read_line();     //call read line
    counter = lsh_cache_line(counter,line, cache);
    printf("This is counter:%i\n", counter);        

    for(i=0; i<10; i++){
      printf("This is cache[%i]:%s\n", i, cache[i]);
    }

    args = lsh_split_line(line);    //split line
    status = lsh_execute(args);     //execute the split args

    free(line);             //free memory
    free(args);
  } while (status);         //continue as long as execute returns 1

}

这是我的程序的输出：

paul@paul-VirtualBox:~/Desktop$ gcc shell.c
paul@paul-VirtualBox:~/Desktop$ ./a.out
/home/paul/Desktop>HI
This is cache[0]:HI
This is counter:1
This is cache[0]:HI
This is cache[1]:(null)
This is cache[2]:(null)
This is cache[3]:(null)
This is cache[4]:(null)
This is cache[5]:(null)
This is cache[6]:(null)
This is cache[7]:(null)
This is cache[8]:(null)
This is cache[9]:(null)
lsh: No such file or directory
/home/paul/Desktop>this is my problem
This is cache[1]:this is my problem
This is counter:2
This is cache[0]:/home/paul/Desktop
This is cache[1]:this is my problem
This is cache[2]:(null)
This is cache[3]:(null)
This is cache[4]:(null)
This is cache[5]:(null)
This is cache[6]:(null)
This is cache[7]:(null)
This is cache[8]:(null)
This is cache[9]:(null)
lsh: No such file or directory
/home/paul/Desktop>it overwrites my previous string with the cwd
This is cache[2]:it overwrites my previous string with the cwd
This is counter:3
This is cache[0]:/home/paul/Desktop
This is cache[1]:/home/paul/Desktop
This is cache[2]:it overwrites my previous string with the cwd
This is cache[3]:(null)
This is cache[4]:(null)
This is cache[5]:(null)
This is cache[6]:(null)
This is cache[7]:(null)
This is cache[8]:(null)
This is cache[9]:(null)
lsh: No such file or directory
/home/paul/Desktop>^C
paul@paul-VirtualBox:~/Desktop$ 

我已经尝试过声明和初始化字符串数组的不同方法，但这种方法似乎效果最好

---

我做错了什么？

缓存中的字符串没有存储空间

类似于

strdup

的东西将创建存储，但稍后需要

释放内存

int lsh_cache_line(int counter,char *line, char *cache[10]){

 (cache[counter]) = strdup(line);
 printf("This is cache[%i]:%s\n", counter, cache[counter]);
 counter++;
 counter = counter % 10;
 return counter; 

}

字符串有10个插槽，但字符串值没有内存。您需要分配一些内存。
我敢打赌，您的lsh\u read\u line（）函数始终会为您返回相同的缓冲区和不同的文本。然后将指向该缓冲区的指针存储到数组中的不同单元格中。一旦将文本放入行变量中，就应该将其复制到新分配的字符串中，以后只使用此新副本。
可能是一些未定义的行为。使用所有警告和调试信息编译（
gcc-Wall-Wextra-g
）。然后使用调试器（
gdb
）和它的监视点；strcpy（tmp，生产线）；高速缓存[计数器]=tmp；到缓存函数，得到了相同的结果，那么char*tmp=malloc（strlen（line）*sizeof（char））呢；strcpy（tmp，生产线）；高速缓存[计数器]=tmp；我不确定你的意思，我用char*cache[10]={NULL}初始化了主循环中的缓存
int lsh_cache_line(int counter,char *line, char *cache[10]){

 (cache[counter]) = strdup(line);
 printf("This is cache[%i]:%s\n", counter, cache[counter]);
 counter++;
 counter = counter % 10;
 return counter; 

}