复制C字符串的正确方法 [英] Proper way to copy C strings
问题描述
是否有一种简单的方法来复制C字符串？
Is there an easy way to copy C strings?
我有 const char * stringA ，我想要 char * stringB 取值（请注意， stringB 不是 const ）。我试过 stringB =（char *）stringA ，但是这使得 stringB 仍然指向相同的内存位置，所以当 stringA 之后会更改， stringB 也会更改。
I have const char *stringA, and I want char *stringB to take the value (note that stringB is not const). I tried stringB=(char*) stringA, but that makes stringB still point to the same memory location, so when stringA later changes, stringB does too.
I还尝试了 strcpy（stringB，stringA），但似乎如果 stringB 没有初始化为大足够的数组，有一个段错误。我对C字符串不是很有经验，我是否缺少明显的东西？
I've also tried strcpy(stringB,stringA), but it seems that if stringB wasn't initialized to a large enough array, there's a segfault. I'm not super experienced with C strings though, am I missing something obvious?
如果我只是初始化 stringB 作为 char * stringB [23] ，因为我知道我的字符串永远不会超过 22 个字符（并且允许使用空终止符），这是正确的方法吗？如果检查 stringB 与其他C字符串是否相等，多余的空间会影响什么吗？
If I just initialize stringB as char *stringB[23], because I know I'll never have a string longer than 22 characters (and allowing for the null terminator), is that the right way? If stringB is checked for equality with other C-strings, will the extra space affect anything?
（在这里使用字符串不是解决方案，因为我需要的开销最小并且可以轻松访问单个字符。）
(And just using strings isn't a solution here, as I need minimal overhead and easy access to individual characters.)
推荐答案
您可以使用 strdup（） 以返回C字符串的副本，例如：
You could use strdup() to return a copy of a C-string, as in:
#include <string.h>
const char *stringA = "foo";
char *stringB = NULL;
stringB = strdup(stringA);
/* ... */
free(stringB);
您还可以使用 strcpy（） ，但是您需要先分配空间，这并不难但如果执行不正确，则会导致溢出错误：
You could also use strcpy(), but you need to allocate space first, which isn't hard to do but can lead to an overflow error, if not done correctly:
#include <string.h>
const char *stringA = "foo";
char *stringB = NULL;
/* you must add one to cover the byte needed for the terminating null character */
stringB = (char *) malloc( strlen(stringA) + 1 );
strcpy( stringB, stringA );
/* ... */
free(stringB);
如果您不能使用 strdup（），我建议使用 strncpy（） 而不是 strcpy（）。 strncpy（）函数最多可复制-且最多可复制- n 个字节，这有助于避免溢出错误。如果 strlen（stringA）+ 1> n ，但是，您需要自己终止 stringB 。但是，通常来说，您会知道自己需要什么大小的东西：
If you cannot use strdup(), I would recommend the use of strncpy() instead of strcpy(). The strncpy() function copies up to — and only up to — n bytes, which helps avoid overflow errors. If strlen(stringA) + 1 > n, however, you would need to terminate stringB, yourself. But, generally, you'll know what sizes you need for things:
#include <string.h>
const char *stringA = "foo";
char *stringB = NULL;
/* you must add one to cover the byte needed for the terminating null character */
stringB = (char *) malloc( strlen(stringA) + 1 );
strncpy( stringB, stringA, strlen(stringA) + 1 );
/* ... */
free(stringB);
我认为 strdup（）更干净，我自己，所以我尝试在专门用于字符串的地方使用它。在性能方面，我不知道POSIX /非POSIX方法是否存在严重的缺点，但我不是C或C ++专家。
I think strdup() is cleaner, myself, so I try to use it where working with strings exclusively. I don't know if there are serious downsides to the POSIX/non-POSIX approach, performance-wise, but I am not a C or C++ expert.
请注意，我将 malloc（）的结果转换为 char * 。这是因为您的问题被标记为 c ++ 问题。在C ++中，需要从 malloc（）强制转换结果。但是，在C语言中，您不强制转换。
Note that I cast the result of malloc() to char *. This is because your question is tagged as a c++ question. In C++, it is required to cast the result from malloc(). In C, however, you would not cast this.
EDIT
你去了，有一个复杂的问题： strdup（）不在C或C ++中。因此，将 strcpy（）或 strncp（）与预大小数组或 malloc一起使用 -ed指针。在任何可能使用该功能的地方，使用 strncp（）而不是 strcpy（）是一个好习惯。这将有助于减少潜在的错误。
There you go, there's one complication: strdup() is not in C or C++. So use strcpy() or strncp() with a pre-sized array or a malloc-ed pointer. It's a good habit to use strncp() instead of strcpy(), wherever you might use that function. It will help reduce the potential for errors.
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