为什么从字符串常量到 'char*' 的转换在 C 中有效但在 C++ 中无效
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Why is conversion from string constant to 'char*' valid in C but invalid in C++
提问by rullof
The C++11 Standard (ISO/IEC 14882:2011) says in § C.1.1:
C++11 标准 (ISO/IEC 14882:2011) 在§ C.1.1：
char* p = "abc"; // valid in C, invalid in C++
For the C++ it's OK as a pointer to a String Literal is harmful since any attempt to modify it leads to a crash. But why is it valid in C?
对于 C++ 来说，它是可以的，因为指向字符串文字的指针是有害的，因为任何修改它的尝试都会导致崩溃。但为什么它在 C 中有效？
The C++11 says also:
C++11 还说：
char* p = (char*)"abc"; // OK: cast added
Which means that if a cast is added to the first statement it becomes valid.
这意味着如果将强制转换添加到第一个语句中，它就会变得有效。
Why does the casting makes the second statement valid in C++ and how is it different from the first one? Isn't it still harmful? If it's the case, why did the standard said that it's OK?
为什么强制转换使第二个语句在 C++ 中有效，它与第一个语句有何不同？不是还是有害的吗？如果是这样，为什么标准说可以？
回答by Jerry Coffin
Up through C++03, your first example was valid, but used a deprecated implicit conversion--a string literal should be treated as being of type char const *, since you can't modify its contents (without causing undefined behavior).
在 C++03 之前，您的第一个示例是有效的，但使用了不推荐使用的隐式转换——字符串文字应该被视为 type char const *，因为您无法修改其内容（不会导致未定义的行为）。
As of C++11, the implicit conversion that had been deprecated was officially removed, so code that depends on it (like your first example) should no longer compile.
从 C++11 开始，已被弃用的隐式转换已被正式删除，因此依赖于它的代码（如您的第一个示例）不应再编译。
You've noted one way to allow the code to compile: although the implicit conversion has been removed, an explicitconversion still works, so you can add a cast. I would not, however, consider this "fixing" the code.
您已经注意到允许代码编译的一种方法：虽然隐式转换已被删除，但显式转换仍然有效，因此您可以添加强制转换。我不是，但是，考虑这个“固定”的代码。
Truly fixing the code requires changing the type of the pointer to the correct type:
真正修复代码需要将指针的类型更改为正确的类型：
char const *p = "abc"; // valid and safe in either C or C++.
As to why it was allowed in C++ (and still is in C): simply because there's a lot of existing code that depends on that implicit conversion, and breaking that code (at least without some official warning) apparently seemed to the standard committees like a bad idea.
至于为什么它在 C++ 中被允许（并且仍然在 C 中）：仅仅因为有很多现有的代码依赖于这种隐式转换，并且破坏该代码（至少没有一些官方警告）显然对标准委员会来说似乎是一个坏主意。
回答by Barmar
It's valid in C for historical reasons. C traditionally specified that the type of a string literal was char *rather than const char *, although it qualified it by saying that you're not actually allowed to modify it.
由于历史原因，它在 C 中有效。C 传统上指定字符串文字的类型是char *而不是const char *，尽管它通过说您实际上不允许修改它来限定它。
When you use a cast, you're essentially telling the compiler that you know better than the default type matching rules, and it makes the assignment OK.
当您使用强制转换时，您实际上是在告诉编译器您比默认类型匹配规则更了解，并且它使分配正常。
回答by baz
You can also use strdup:
您还可以使用strdup：
char* p = strdup("abc");
