如何使一个不是空终止的C字符串？ [英] how to make a not null-terminated c string?

问题描述

我想知道：字符* CS = .....;会发生什么的strlen（）和printf（％S，CS）如果CS点内存块是巨大的，但没有'\\ 0'在里面？
 我写这几行：

 字符S2 [3] = {'A​​'，'A'，'A'};
的printf（str是％S，长度为％D，S2，strlen的（S2））;
 

我得到的结果是：AAA，3，但我认为这样的结果是因为一个'\\ 0'（或0字节）恰好位于该位置S2 + 3。
如何使一个不是空终止的C字符串？ strlen的和其他C字符串函数在很大程度上依赖于'\\ 0'字节，如果没有'\\ 0'，我只是想知道这条规则更深，更好。

PS：我的好奇心是通过研究所以follw后引起。
如何转换一个const char *到std :: string的
而这些字在该职位：
这实际上是麻烦比它的外观，因为除非该字符串实际上是NUL结尾你不能调用strlen的。

解决方案

如果这不是空终止，那么它是不是C字符串，而不能使用的功能，如的strlen - 他们会扬长而去数组的末尾，从而导致未定义的行为。你需要保持长的赛道一些其他的方式。

您仍然可以打印非终止字符数组与的printf ，只要你给的长度：

 的printf（str是％.3s，S2）;
的printf（str是％* S，s2_length，S2）;
 

或者，如果你有机会到阵列本身，而不是一个指针：

 的printf（str是％* S。（INT）（sizeof的S2），S2）;
 

您也已经标记的问题C ++：在语言中，你通常要避免这一切容易出错说大话，并使用的std ::字符串而不是

i am wondering :char *cs = .....;what will happen to strlen() and printf("%s",cs) if cs point to memory block which is huge but with no '\0' in it? i write these lines:

 char s2[3] = {'a','a','a'};
printf("str is %s,length is %d",s2,strlen(s2));

i get the result :"aaa","3",but i think this result is because that a '\0'(or a 0 byte) happens to reside in the location s2+3. how to make a not null-terminated c string? strlen and other c string function relies heavily on the '\0' byte,what if there is no '\0',i just want know this rule deeper and better.

ps: my curiosity is aroused by studying the follw post on SO. How to convert a const char * to std::string and these word in that post : "This is actually trickier than it looks, because you can't call strlen unless the string is actually nul terminated."

解决方案

If it's not null-terminated, then it's not a C string, and you can't use functions like strlen - they will march off the end of the array, causing undefined behaviour. You'll need to keep track of the length some other way.

You can still print a non-terminated character array with printf, as long as you give the length:

printf("str is %.3s",s2);
printf("str is %.*s",s2_length,s2);

or, if you have access to the array itself, not a pointer:

printf("str is %.*s", (int)(sizeof s2), s2);

You've also tagged the question C++: in that language, you usually want to avoid all this error-prone malarkey and use std::string instead.

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