C语言 获取字符串的前 10 个字符？

Getting the first 10 characters of a string?

提问by user2341069

I have not been able to find any information with a web-search. Where should I be looking?

我无法通过网络搜索找到任何信息。我应该去哪里找？

回答by Lefteris E

char myString[256]; // Input string
char dest[256];     // Destination string

strncpy(dest, myString, 10);
dest[10] = 0; // null terminate destination

回答by tianz

char source[] = "abcdefthijklmn";
char target[100];

strncpy(target, source, 10);
target[10] = '
char* someString = "your string goes here";

int main() 
{
  int n = 10;
  printf("(%.*s)\n", n, someString);

  return 0;
}
'; // IMPORTANT!

回答by Ashish Srivastava

Adding to the above answers:

补充上面的答案：

char source[] = "abcdefthijklmn";
char target[100];

snprintf(target, sizeof target, "%.10s", source);

回答by chqrlie

There are many different ways to achieve your goal:

有许多不同的方法可以实现您的目标：

• You can use snprintf(safest):

  char source[] = "abcdefthijklmn";
  char target[100];
  
  *target = '
  char source[] = "abcdefthijklmn";
  char target[100];
  size_t len = strlen(source);
  
  if (len > 10)
      len = 10;
  memcpy(target, source, len);
  target[len] = '
  char source[] = "abcdefthijklmn";
  char target[100];
  size_t i;
  for (i = 0; i < 10; i++) {
      target[i] = source[i];
  }
  target[i] = '
  char source[] = "abcdefthijklmn";
  char target[100];
  
  snprintf(target, sizeof target, "%.10s", source);
  ';
  ';
  ';
  strncat(target, source, 10);
  

• You can use strncatif you know the destination has at least 11 elements:

  char source[] = "abcdefthijklmn";
  char target[100];
  
  *target = '
  char source[] = "abcdefthijklmn";
  char target[100];
  size_t len = strlen(source);
  
  if (len > 10)
      len = 10;
  memcpy(target, source, len);
  target[len] = '
  char source[] = "abcdefthijklmn";
  char target[100];
  size_t i;
  for (i = 0; i < 10; i++) {
      target[i] = source[i];
  }
  target[i] = '
     # include <stdio.h>
     # include <string.h>
     //Strings. String lenght terminator.
     //KHO2016.no1. mingw (TDM-GCC-32) . c-ansi .
     int main ()
     {
     //declare
     char src_str[20],dst_str[10];
  
     //valuate
     printf ("Enter a sentance of 20 letters\n");
     gets (src_str);
     strcpy (dst_str,src_str);
  
     //calculate
     dst_str [10] ='
  #include <stdio.h>
  #include <string.h>
  
  int main(void)
  {   
      char source[] = "abcdefghijklmnopqrstuvwxyz";
      char dest[11];
      memset(dest, '##代码##', sizeof(dest));
  
      sprintf(dest, "%.10s", source);
  
      printf("%s", dest); // abcdefghij
      return 0;
  }
  '; // from the "www.stack overflow"
     printf ("%s",dst_str);
     printf ("\n");
  
     //terminate
     return 0;
     }
  ';
  ';
  ';
  strncat(target, source, 10);
  

• You can use strlenand memcpy(same assumption about the size of destination):

  ##代码##

• You can use a loop (same assumption about the size of destination):

  ##代码##

• You could but should notuse strncpy()

• 您可以使用snprintf（最安全）：

  ##代码##

• strncat如果您知道目的地至少有 11 个元素，则可以使用：

  ##代码##

• 您可以使用strlen和memcpy（关于目的地大小的相同假设）：

  ##代码##

• 您可以使用循环（对目的地大小的相同假设）：

  ##代码##

• 你可以但不应该使用strncpy()

回答by TonyArra

If you're looking for a good source, here's an example of an online man page you can use: http://linux.die.net/man/3/strncpy

如果您正在寻找一个好的来源，这里有一个您可以使用的在线手册页示例：http: //linux.die.net/man/3/strncpy

Important to note: you could also use memcpy instead of strncpy, but it requires you to add your own null-terminating byte.

需要注意的重要事项：您也可以使用 memcpy 代替 strncpy，但它要求您添加自己的空终止字节。

Warning: If there is no null byte among the first n bytes of src, the string placed in dest will not be null-terminated.

警告：如果 src 的前 n 个字节中没有空字节，则放置在 dest 中的字符串不会以空值结尾。

Hence, memcpy and strncpy work almost the same here, and memcpy is more efficient and less prone to error.

因此，memcpy 和 strncpy 在这里的工作方式几乎相同，并且 memcpy 更高效且不易出错。

回答by Radar Blue

I got it to work like this.

我让它像这样工作。

##代码##

回答by w.b

You can also use sprintfwith .10precision format:

您还可以使用sprintf与.10精度格式：

##代码##