Question

我有一个字符串数组，我想在用户输入字符串时动态更改。此2D数组具有固定大小，例如char A[N][BUFF_MAX]将存储N个字符串，每个字符串最多为BUFF_MAX。我希望这个数组循环，也就是说，最新的输入应该存储在A[0]中，依此类推。

到目前为止，我在实现上述方面没有任何问题。当用户输入超过N个字符串时，我的问题就出现了。此时我想存储最后 N个输入，并将其余输出丢弃为仅使用N个单元格。可以使用strcpy进行移位。

现在，我想用字符串编号打印这个数组。例如，如果用户输入M字符串，其中M < N字符串，则应打印

    1  string_1
    2  string_2
    :  :
    M  string_M

但是，如果M > N则应打印

  M-N  string_(M-N)
M-N+1  string_(M-N+1)
    :  :
    M  string_M

我尝试了几件事情，但我无法使其发挥作用。这是我的代码的简化版本。

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

#define N 10                                      /* max size of array */
#define BUFF_MAX 50                               /* max size of buffer */
int M = 0;                                        /* input count */
char A[N][BUFF_MAX];                              /* should i initialize this? */

void displayArray() {                        
    for (int h = 0; h < M; h++) {                 /* print the elements */
        printf("%2d  %s\n", h + 1, A[h]);         /* this doesn't number properly */
                                                  /* and needs to be modified */
    }
}

int main(void) {
    while(1) {                                    /* keep reading from the user */
        char input[BUFF_MAX];
        fgets(input, sizeof(input), stdin);       /* read input from user */
        strcpy(A[M], input);                      /* copy input into array */
        M++;                                      /* i might need some modulo here */

        if (M >= N) {                             /* if user inputs more than N */
            for (int i = N - 1; i > 0; i--) {     /* shift the array */
                strcpy(A[i], A[i - 1]); 
            }
        }
        if (!strcmp(input, "hist"))
            displayArray();
    }    
}

感谢您的帮助。

Answer 1

这是您的代码，但有一些更改

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

#define N 10                                      /* max size of array */
#define BUFF_MAX 50                               /* max size of buffer */
int M = 0;                                        /* input count */
char A[N][BUFF_MAX];                              /* should i initialize this? */

void displayArray()
{
    int h;
    //you need to change the condition of for in order to print
    // the adequate number of elements
    for ( h = 0; h < (M>N?N:M); h++)                   /* print the elements */
    {
        printf("%2d  %s\n", h + 1, A[h]);         /* this doesn't number properly */
        /* and needs to be modified */
    }
}

int main(void)
{
    while(1)                                      /* keep reading from the user */
    {
        char input[BUFF_MAX];
        fgets(input, sizeof(input), stdin); /* read input from user */
        // you need to add the condition below because the fgets include the newline
        // if the user hit enter after each input
        if(input[strlen(input)-1]=='\n')
            input[strlen(input)-1]='\0';
        // note that the subscript of the array is changed in order to
        // get the desired input A[ M % N ]
        // so the sequence of index will be 0 1 2 ... N-1 0 1 2 etc.
        strcpy(A[M % N], input);                      /* copy input into array */
        M++;                                      /* i might need some modulo here */

        if (!strcmp(input, "hist"))
            displayArray();
    }

    return 0;
}

请记住，您的代码没有断点才能退出程序。您需要在break循环中添加return或goto或while才能完成执行!!

Answer 2

这就是我的想法。请注意，我避免“无限循环”并且程序检测到EOF并在EOF时停止。输入hist会触发打印操作，但该行不会保存在历史记录中。此外，如果你观察，你会注意到10行缓冲区中只保存了9个条目。如果您必须显示10行，请将N增加1。

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

#define N 10          /* max size of array */
#define BUFF_MAX 50   /* max size of buffer */

int M = 0;            /* monotonic input count */
int first_valid = 0;
int first_avail = 0;
char A[N][BUFF_MAX];

static
void displayArray(int v1, int v2)
{
    int h = v1;
    int x = M - N + 1;
    if (x < 0)
        x = 0;
    while (h != v2)
    {
        printf("%2d  %s\n", ++x, A[h++]);
        if (h >= N)
            h = 0;
    }
}

int main(void)
{
    char input[BUFF_MAX];
    while (fgets(input, sizeof(input), stdin) != 0)
    {
        int len = strlen(input);
        if (input[len-1] == '\n')
            input[--len] = '\0';
        if (strcmp(input, "hist") == 0)
            displayArray(first_valid, first_avail);
        else
        {
            strcpy(A[first_avail++], input);
            M++;
            if (first_avail >= N)
                first_avail = 0;
            if (first_avail == first_valid)
            {
                if (++first_valid >= N)
                    first_valid = 0;
            }
        }
    }
    return 0;
}

示例运行：

Line 1
hist
 1  Line 1
line 2
hist
 1  Line 1
 2  line 2
line 3
xyz 4
asd 5
hist
 1  Line 1
 2  line 2
 3  line 3
 4  xyz 4
 5  asd 5
Emergency 6
Helpline 7
hist
 1  Line 1
 2  line 2
 3  line 3
 4  xyz 4
 5  asd 5
 6  Emergency 6
 7  Helpline 7
Looping Later 8
Last Line Before Recycling 9
hist
 1  Line 1
 2  line 2
 3  line 3
 4  xyz 4
 5  asd 5
 6  Emergency 6
 7  Helpline 7
 8  Looping Later 8
 9  Last Line Before Recycling 9
First lot of recycling 10
hist
 2  line 2
 3  line 3
 4  xyz 4
 5  asd 5
 6  Emergency 6
 7  Helpline 7
 8  Looping Later 8
 9  Last Line Before Recycling 9
10  First lot of recycling 10
Keep going
Dozens of lines
Bad Luck
A fortnight
hist
 6  Emergency 6
 7  Helpline 7
 8  Looping Later 8
 9  Last Line Before Recycling 9
10  First lot of recycling 10
11  Keep going
12  Dozens of lines
13  Bad Luck
14  A fortnight

使用全局计数器循环通过C数组

2 个答案: