Question

我想将各种大小的字符串放入2D数组中，表示为char *array[size];各种字符串大小的示例可以是： - "Hi", "Welcome";。

这应该适合上面给出的数组：如果数组以这种格式给出：char array[][]我可以使用下面的代码来完成它。：

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

char array[][100];

int main()
{    
    char buf[10];

    sprintf(buf,"%d", 12);
    strcpy(array[0], buf);
    sprintf(buf, "%s", "hello");
    strcpy(array[1], buf);
    printf("%s %s", array[0], array[1]);
}

但我无法做到，如果数组以char *array[]格式给出。

任何帮助都会非常有帮助。谢谢。

Answer 1

由于你已经对这两个字符串进行了硬编码，你可以像这样声明一个指针数组：

char *array[] = {"Hi", "Welcome"};

这些字符串文字是只读的。更详尽的方法可能是这样的：

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

#define STRINGS 2
#define MAXLEN  100

int main(void)
{    
    char buf[MAXLEN];
    char *array[STRINGS];                       // array of string pointers
    int i;
    for (i=0; i<STRINGS; i++) {
        printf("Enter string %d: ", i+1);
        if (fgets(buf, MAXLEN, stdin) == NULL)
            return 0;                           // bad input
        buf [ strcspn(buf, "\r\n") ] = 0;       // remove trailing newline etc
        array[i] = malloc(strlen(buf)+1);       // only as much as needed
        if (array[i] == NULL)
            return 0;                           // bad memory allocation
        strcpy(array[i], buf);
    }

    // print the array
    for (i=0; i<STRINGS; i++) {
        printf("%s\n", array[i]);
    }

    // free the memory
    for (i=0; i<STRINGS; i++) {
        free(array[i]);
    }
    return 0;
}

Answer 2

在C中有两种不同类型的数据结构称为“二维数组”。由于运算符重载，两者都有table[i]处理的行和table[i][j]处理的元素。

逻辑上细分为行大小的块的连续内存块有时称为矩形数组（因为如果将元素放置为表格，则所有行都将完美排列）并且一个指针数组有时被称为不规则数组。如果要使用其中一个，则需要自己为每一行分配内存。这是一些混合两者的示例代码。

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

#define ROWS 100U
#define COLS 32U

char rectangular_array[ROWS][COLS] = {'\0'};
char *ragged_array[ROWS] = {NULL};

int main(void) {
  strncpy( rectangular_array[0], "hello,", COLS-1 );
  strncpy( rectangular_array[1], "world!", COLS-1 );
  /* I don't need to set the third string to "" or the final bytes to
   * '\0' explicitly because I initialized the array to zeroes.
   */

  size_t i = 0;
  while( i < ROWS && rectangular_array[i][0] ) {
    const char* const s = &rectangular_array[i][0];
    const size_t m = strlen(s)+1;
    char* const t = malloc(m);

    printf( "%s ", s );
    // Copy the contents into ragged_array.
    assert(t);
    memcpy( t, s, m );
    ragged_array[i] = t;
    ++i;
  }

  if ( i < ROWS )
    ragged_array[i] = NULL;
  /* We DO, however, need to set the termiating entry of this array to
   * NULL, because, contrary to common misconception, a pointer with all
   * bits zeroed out is not necessarily a NULL pointer, and there are
   * implementations in the real world where NULL is a different
   * special value that traps on some hardware.
   *
   * Confusingly, the standard does say that assigning the constant 0 to
   * a pointer sets it to NULL.
   */

  puts("\n");

  for ( i = 0; i < ROWS && ragged_array[i]; ++i ) {
    // Print the contents of each row of ragged_array, then free each row.
    printf( "%s ", ragged_array[i] );
    free(ragged_array[i]);
    ragged_array[i] = NULL;
  }

  puts("\n");

  return EXIT_SUCCESS;
}

如何将各种长度的字符串放在指针中，比如char * output [100]

2 个答案: