C语言 如何在c中获取char数组的一部分
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how to get a part of a char array in c
提问by dali1985
I am new in C and I have the following simple code. I know that with the strncpy I can copy characters from string.
我是 C 新手，我有以下简单的代码。我知道使用 strncpy 我可以从字符串中复制字符。
#include <stdio.h>
#include <string.h>
int main ()
{
char str1[]= "To be or not to be";
char str2[40];
strncpy ( str2, str1, 5 );
str2[5] = 'To be
'; /* null character manually added */
puts (str2);
return 0;
}
The output of this code is
这段代码的输出是
strncpy ( str2, str1+6, 9);
str2[9] = 'strncpy ( str2, str1+strlen("To be "), strlen("or not to") );
str2[strlen("or not to")] = 'strncpy ( str2, str1 + 6, 9 );
';
'; /* null character manually added */
If I want the result to be ′or not to′ how can I read these characters? From 7-15 in this case?
如果我希望结果是“或不”，我该如何读取这些字符？从 7-15 在这种情况下？
回答by Vijay
Use :
用 ：
##代码##回答by Dayal rai
No need to count manually ,you can use standard library function for this.Try like following
无需手动计数，您可以为此使用标准库函数。尝试如下
##代码##回答by Bart Friederichs
Simple way to do that, offset str1:
这样做的简单方法，偏移量str1：
