C++ 将两个 const char* 组合在一起
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Combining two const char* together
提问by 111WARLOCK111
I've tried many ways to do that, I got a void which is static and is on Console class i made, Void it self works fine:
我已经尝试了很多方法来做到这一点，我得到了一个静态的 void，它在我制作的 Console 类上，Void 本身可以正常工作：
Console::WriteLine(const char* msg)
On the other side, I got another const char* non static void which calls the Console::WriteLine void from It, I've been working on C# for around a year and on C# I can easily do something like this:
另一方面，我得到了另一个 const char* non static void，它从中调用了 Console::WriteLine void，我已经在 C# 上工作了大约一年，在 C# 上我可以轻松地做这样的事情：
string a = "Start ";
string b = a + "End";
When i call this on C++, it gives me bunch of errors:
当我在 C++ 上调用它时，它给了我一堆错误：
Player::Kill(const char* Message)
{
Console::WriteLine("> " + Message + " <");
}
I've also tried the strcat thing and put, But it tells me to use strcat_s which doesn't really work, And also I've tried to do string instead of const char*, And tried char*, But all of them give errors for the thing I'm trying to do.
我也尝试过 strcat 的东西并放置，但它告诉我使用 strcat_s 这并不真正有效，而且我尝试使用 string 而不是 const char*，并尝试使用 char*，但所有这些都给出我正在尝试做的事情的错误。
回答by kfsone
"const" means "cannot be changed(*1)". So you cannot simply "add" one const char string to another (*2). What you can do is copy them into a non-const character buffer.
“const”的意思是“不能改变(*1)”。因此，您不能简单地将一个 const 字符字符串“添加”到另一个 (*2) 中。您可以做的是将它们复制到非常量字符缓冲区中。
const char* a = ...;
const char* b = ...;
char buffer[256]; // <- danger, only storage for 256 characters.
strncpy(buffer, a, sizeof(buffer));
strncat(buffer, b, sizeof(buffer));
// now buffer has the two strings joined together.
Your attempt to use std::string failed for a similar reason. You said:
由于类似的原因，您尝试使用 std::string 失败。你说：
std::string a = "Start";
std::string b = a + " End";
This translates to
这转化为
b = (std::string)a + (const char*)" End";
Which should be ok except that it creates an extra string, what you probably wanted is
这应该没问题，除了它会创建一个额外的字符串，你可能想要的是
std::string a = "Start";
a += " End";
If you are getting compile errors doing this, please post them (Make sure you #include ).
如果您在执行此操作时遇到编译错误，请发布它们（确保您使用 #include ）。
Or you could do something like:
或者你可以这样做：
std::string addTwoStrings(const std::string& a, const std::string& b)
{
return a + b; // works because they are both strings.
}
All of the following work: (see live demo http://ideone.com/Ytohgs)
以下所有工作：（见现场演示http://ideone.com/Ytohgs）
#include <iostream>
#include <string>
std::string addTwoStrings(const std::string& a, const std::string& b)
{
return a + b; // works because they are both strings.
}
void foo(const char* a, const char* b)
{
std::string str = a;
std::cout << "1st str = [" << str << "]" << std::endl;
str += " ";
std::cout << "2nd str = [" << str << "]" << std::endl;
str += b;
std::cout << "3rd str = [" << str << "]" << std::endl;
str = addTwoStrings(a, " ");
std::cout << "4th str = [" << str << "]" << std::endl;
str = addTwoStrings(str, b);
std::cout << "5th str = [" << str << "]" << std::endl;
}
int main()
{
foo("hello", "world");
}
*1 Or more accurately, "cannot be changed in-situ" - you can use it in expressions, etc, so for example, e.g.
*1 或更准确地说，“无法原位更改”-您可以在表达式等中使用它，例如，例如
const size_t len = strlen("hello");
size_t newLen = len + strlen("world");
// but this would not be legal:
len += 2; // error: len is const.
2 "const chara + const char* b" is actually trying to add two pointersnot two strings, the result would be the address of string a plus the address of string b, the sum of which would be some random memory location
2 “const chara + const char* b”实际上是尝试添加两个指针而不是两个字符串，结果将是字符串 a 的地址加上字符串 b 的地址，其总和将是某个随机内存位置
回答by Alexis Wilke
char *is a pointer (so are "> "and " <"), you cannot add pointers together.
char *是一个指针（"> "和也是" <"），你不能把指针加在一起。
However you can concatenate C++ strings using the + operator:
但是，您可以使用 + 运算符连接 C++ 字符串：
Player::Kill(const std::string& Message)
{
Console::WriteLine(("> " + Message + " <").c_str());
}
回答by JaredPar
Instead of concatenating the strings and creating an extra temporary object, why not just output the 3 strings separately?
与其连接字符串并创建一个额外的临时对象，为什么不单独输出 3 个字符串？
Player::Kill(const char* Message)
{
Console::Write("> ");
Console::Write(Message);
Console::WriteLine(" <");
}
回答by Dietmar Kühl
Since you say it's C++ code, just just this:
既然你说它是 C++ 代码，就这样：
void Player::Kill(std::string const& Message)
{
Console::WriteLine(("> " + Message + " <").c_str());
}
Ideally, your Console::WriteLine()is declared to also take a std::string const&in which case you don't need to do the .c_str()-dance.
理想情况下，你Console::WriteLine()被声明为也采取 astd::string const&在这种情况下你不需要做.c_str()-dance。
