为什么我可以在 C 中将 int 文字隐式转换为 int * 而在 C++ 中不能? [英] Why can I implicitly convert an int literal to an int * in C but not in C++?

问题描述

我相信在下面的代码中，C自动将 17 转换为 int *"，正如最近有人指出的(但没有给出原因)，这是错误的.

I believed that in the following code, C "automatically casts 17 to an int *" which, as someone recently pointed out (but did not give the reasons as to why), is wrong.

int *ptoi = 17; // I assumed that 17 is being automatically casted to int *

我知道如果我在 C++ 中做与上面相同的事情，我会收到一个错误消息，说 invalid conversion from int to int *.但是，如果我在 C++ 中执行以下操作，则效果很好:

I know that if I do the same thing as above in C++, I get an error saying invalid conversion from int to int *. But if I do the following in C++, it works fine:

int *ptoi = (int *)17;

这些是我认为在 C 中强制转换是隐式的原因.

These are the reasons I thought that in C, the casting was implicit.

谁能解释一下为什么在 C++ 中，我必须强制转换它，而在 C 中，它可以正常工作?

Can someone please explain why, in C++, I have to cast it but in C, it works fine?

推荐答案

在 C 中，从整数到指针的转换在没有强制转换的情况下也是非法的.不过，大多数编译器会让你摆脱它.Clang 发出警告:

Conversions from integers to pointers without casts are also illegal in C. Most compilers will let you get away with it though. Clang gives a warning:

example.c:5:8: warning: incompatible integer to pointer conversion initializing
      'int *' with an expression of type 'int'
  int *x = 17;
       ^   ~~

C99 在第 6.5.4 节强制转换运算符，第 4 段中说:

C99 says in Section 6.5.4 Cast operators, paragraph 4:

涉及指针的转换，除非 6.5.16.1 的约束允许，应通过显式转换来指定.

Conversions that involve pointers, other than where permitted by the constraints of 6.5.16.1, shall be specified by means of an explicit cast.

6.5.16.1 是 void * 无需强制转换即可转换为其他指针的例外.

6.5.16.1 is the exception for void * converting to other pointers without needing a cast.

C++ 规范在第 5.4 节显式类型转换(强制转换符号)，第 3 段中说:

The C++ spec says in Section 5.4 Explicit type conversion (cast notation), paragraph 3:

以下未提及且未由用户明确定义的任何类型转换都是格式错误的.

Any type conversion not mentioned below and not explicitly defined by the user is ill-formed.

所以你去 - 在这两种语言中都是非法的，但是为了与许多旧软件的兼容性，很多 C 编译器会让你逃脱它.

So there you go - illegal in both languages, but for compatibility with lots of older software, a lot of C compilers will let you get away with it.

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