这与null终止符无关. 字符串 必须以空值结尾.

This has nothing to do with the null terminator. a string must be null-terminated.

您在此处遇到尾随换行符(\n)的问题.您必须先剥离该换行符，然后再将字符串传递给printf().

You're facing issues with the trailing newline (\n) here. you have to strip off that newline before passing the string to printf().

最简单的方法[需要修改str "]:您可以使用

Easiest way [requires modification of str]: You can do this with strcspn(). Pseudo code:

str[strcspn(str,"\n")] = 0;

(如果可能)在不修改字符串的情况下实现此输出.

是的，也是可能的.在这种情况下，您需要在printf()中使用length修饰符来限制要打印的数组的长度，例如

Yes, possible, too. In that case, you need to use the length modifier with printf() to limit the length of the array to be printed, something like,

printf("%15s", str);  //counting the ending `.` in str as shown

但是恕我直言，这不是***的方法，因为必须知道并固定字符串的长度，否则它将无法正常工作.

but IMHO, this is not the best way, as, the length of the string has to be known and fixed, otherwise, it won't work.

一种灵活的情况，

printf("%.*s", n, str);

其中，必须提供n，并且它需要保留要打印的字符串的长度(没有换行符)

where, n has to be supplied and it needs to hold the length of the string to be printed, (without the newline)