Question

#include <stdio.h>
#include <string.h>
void main(int ac, char *av[])
{
     char str1[] = "this is a test";
     char str2[20];
      char str3[30];
    strncpy(str2,str1,5);

}

我想从字符串str1的索引0开始将字符串str1的5个字符复制到str2中，然后将字符串str1的5个字符从string1的索引1开始复制到str2，依此类推。例如，第一个str2应为“this”。第二个str2 =“他的我”。第三个str2“是”。请帮帮忙，谢谢

Answer 1

只需将您的偏移量添加到str1调用的strncpy参数即可。例如：

strncpy(str2,str1 + 1,5);

将从索引1开始从str1复制5个字节到str2。

Answer 2

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

int main()
{
     char str1[] = "this is a test";
     char str2[20];
     char str3[30];

     strncpy( str2, str1, 5 );
     str2[5] = '\0';
     strncpy( str3, str1 + 1, 5 );
     str3[5] = '\0';

     //...
}

这是一个更完整的例子

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

int main()
{
    char str1[] = "this is a test";
    char str2[sizeof( str1 ) - 5][6];
    const size_t N = sizeof( str1 ) - 5;
    size_t i;

    for ( i = 0; i < N; i++ )
    {
        strncpy( str2[i], str1 + i, 5 );
        str2[i][5] = '\0';
    }

    for ( i = 0; i < N; i++ )
    {
        puts( str2[i] );
    }

    return 0;
}

输出

this 
his i
is is
s is 
 is a
is a 
s a t
 a te
a tes
 test

Answer 3

您尝试做的事情需要特别注意字符串索引和指针偏移。这并不困难，但如果您尝试读取/写出界限，则会立即输入未定义的行为。下面的示例提供了显示其发生情况的输出，以便您可以直观地看到该过程。

我并不完全清楚您的确切目的或您打算如何使用str3，但无论如何，以下原则都适用：

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

int main()
{
    char str1[] = "this is a test";
    char str2[20] = {0};                        /* always initialize variables  */
    // char str3[30] = {0};

    size_t i = 0; 
    char p = 0;                                 /* temp char for output routine */

    for (i = 0; i < strlen (str1) - 4; i++)     /* loop through all chars in 1  */
    {                                           /* strlen(str1) (- 5 + 1) = - 4 */
        strncpy (str2+i, str1+i, 5);            /* copy from str1[i] to str2[i] */

        /* all code that follows is just for output */
        p = *(str1 + i + 5);                    /* save char at str1[i]         */
        *(str1 + i + 5) = 0;                    /* (temp) null-terminate at i   */
        /* print substring and resulting str2 */
        printf (" str1-copied: '%s' to str2[%zd], str2: '%s'\n", str1+i, i, str2);
        *(str1 + i + 5 ) = p;                   /* restor original char         */        
    }

    return 0;
}

<强>输出：

$ ./bin/strpartcpy
 str1-copied: 'this ' to str2[0], str2: 'this '
 str1-copied: 'his i' to str2[1], str2: 'this i'
 str1-copied: 'is is' to str2[2], str2: 'this is'
 str1-copied: 's is ' to str2[3], str2: 'this is '
 str1-copied: ' is a' to str2[4], str2: 'this is a'
 str1-copied: 'is a ' to str2[5], str2: 'this is a '
 str1-copied: 's a t' to str2[6], str2: 'this is a t'
 str1-copied: ' a te' to str2[7], str2: 'this is a te'
 str1-copied: 'a tes' to str2[8], str2: 'this is a tes'
 str1-copied: ' test' to str2[9], str2: 'this is a test'

如何将部分字符串复制到另一个字符串？

3 个答案: