C++ 如何将 const char* 转换为 char*
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原文地址: http://stackoverflow.com/questions/833034/
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How to convert const char* to char*
提问by Cute
can any body tell me how to conver const char* to char*?
有人能告诉我如何将 const char* 转换为 char* 吗？
get_error_from_header(void *ptr, size_t size, size_t nmemb, void *data) {
ErrorMsg *error = (ErrorMsg *)data;
char* err = strstr((const char *)ptr,"550");
//error cannot convert const char** to char*
if(err) {
strncpy(error->data,(char*)ptr,LENGTH_ERROR_MESSAGE-1);
error->data[LENGTH_ERROR_MESSAGE-1] = 'if (NULL != strstr((const char *)ptr, "550"))
{
';
error->ret = true;
}
return size*nmemb;
}
采纳答案by Daniel Earwicker
You don't appear to use errin the rest of that function, so why bother creating it?
您似乎没有err在该函数的其余部分使用，那为什么还要创建它呢？
const char* err = strstr((const char *)ptr, "550");
If you do need it, will you really need to modify whatever it points to? If not, then declare it as constalso:
如果你确实需要它，你真的需要修改它指向的任何东西吗？如果不是，则将其声明为const：
if (NULL != strstr(reinterpret_cast<const char *>(ptr), "550"))
{
Finally, as casts are such nasty things, it is best to use a specific modern-style cast for the operation you want to perform. In this case:
最后，由于强制转换是如此令人讨厌的事情，因此最好对要执行的操作使用特定的现代风格强制转换。在这种情况下：
char *
strstr(const char *s1, const char *s2);
回答by Jared Oberhaus
There are a few things I don't understand here. I see that this is tagged for C++/CLI, but what I describe below should be the same as Standard C++.
这里有几件事我不明白。我看到这被标记为C++/CLI，但我在下面描述的应该与标准 C++相同。
Doesn't compile
不编译
The code you give doesn't compile; get_error_from_headerdoes not specify a return type. In my experiments I made the return type size_t.
您提供的代码无法编译；get_error_from_header不指定返回类型。在我的实验中，我做了返回类型size_t。
Signature of C++ strstr()
C++的签名 strstr()
The signature for strstr()in the standard C library is:
strstr()标准 C 库中的签名是：
const char * strstr ( const char * str1, const char * str2 );
char * strstr ( char * str1, const char * str2 );
but the signature for strstr()in the C++ library, depending on the overload, is one of:
但对于签名strstr()在C ++库，这取决于过载，是以下之一：
const char* err = strstr((const char *)ptr, "550");
if (err != NULL) {
...
}
I would choose the first overload, because you don't want to modify the string, you only want to read it. Therefore you can change your code to:
我会选择第一个重载，因为你不想修改字符串，你只想读取它。因此，您可以将代码更改为：
//error cannot convert const char** to char*
Also, I'm assuming your comment reporting the error:
另外，我假设您的评论报告了错误：
if (strstr((const char *)ptr, "550") != NULL) {
...
}
is a typo: there's no const char**to be seen.
是一个错字：没有const char**被看到。
Assignment to errunnecessary
分配给err不必要的
As is pointed out in a previous answer, the use of errto store the result of strstris unnecessary if all it's used for is checking NULL. Therefore you could use:
正如在之前的答案中指出的那样，如果使用err来存储 的结果strstr只是检查，则没有必要使用它NULL。因此，您可以使用：
if (strstr(reinterpret_cast<const char *>(ptr), "550") != NULL) {
...
}
Use of reinterpret_cast<>encouraged
reinterpret_cast<>鼓励使用
As is pointed out in another answer you should be using reinterpret_cast<>instead of C-style casts:
正如在另一个答案中指出的那样，您应该使用reinterpret_cast<>而不是 C 风格的强制转换：
const char * p1;
char * p2;
p2 = const_cast<char *>(p1);
Use of const_cast<>to strip const
使用的const_cast<>以条const
Given the example in the question, I don't see where this is necessary, but if you had a variable that you need to strip of const-ness, you should use the const_cast<>operator. As in:
鉴于问题中的示例，我看不出这有什么必要，但是如果您有一个需要const去除 -ness的变量，则应该使用const_cast<>运算符。如：
char* err = strstr((char *)ptr,"550");
As is pointed out in a comment, the reason to use const_cast<>operator is so that the author's intention is clear, and also to make it easy to search for the use of const_cast<>; usually stripping const is the source of bugs or a design flaw.
正如评论中所指出的，使用const_cast<>operator的原因是为了让作者的意图明确，也为了便于搜索使用const_cast<>; 通常剥离 const 是错误或设计缺陷的来源。
回答by Matthew Flaschen
Can't you just do:
你不能只做：
const char* mstr="";
char* str=const_cast<char*>(mstr);
The error is because if you pass in a const char* to strstr you get one out (because of the overload).
错误是因为如果您将 const char* 传递给 strstr ，则会得到一个（因为过载）。
回答by hmehdi
//try this instead:
//试试这个：
##代码##回答by Yuliy
Well is ptr (which you passed in as void*) actually const or not? (In other words, is the memory under your control?) If it's not, then cast it to char* instead of const char* when calling strstr. If it is, however, you'll get a const char* out (pointing to a location inside of the string pointed to by ptr), and will then need to strncpy out to another string which you are responsible for managing.
好吧，ptr（你作为 void* 传入）实际上是不是 const ？（换句话说，内存是否在您的控制之下？）如果不是，则在调用 strstr 时将其转换为 char* 而不是 const char*。但是，如果是这样，您将得到一个 const char* 输出（指向 ptr 指向的字符串内部的位置），然后需要将 strncpy 输出到您负责管理的另一个字符串。
