Why can I implicitly convert an int literal to an int * in C but not in C++?

问题

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 *

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;

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

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

回答1:

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 says in Section 6.5.4 Cast operators, paragraph 4:

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

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

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

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.

回答2:

Yes, the casting is implicit in C, although many (all?) compilers give a warning. In C++, no implicit conversion is performed from int to int*, so the explicit cast is required.

回答3:

It has been, C has implicit conversions - in my experience pretty much from any integral type to any other integral type. Pointers in C are ~~considered integral~~scalar types.

In C++, Integral Types are defined as follows:

Types bool, char, char16_t, char32_t, wchar_t, and the signed and unsigned integer types are collectively called integral types.⁴⁸ A synonym for integral type is integer type. The representations of integral types shall define values by use of a pure binary numeration system.⁴⁹ [ Example: this International Standard permits 2’s complement, 1’s complement and signed magnitude representations for integral types. —end example ]

The only integral value that can be converted to pointer type in C++ is the null pointer constant, though technically the conversion is to a prvalue of type std::nullptr_t. (para 4.10)

Final Note

Instead of fixing it like this:

int *ptoi = (int *)17;

considering adding the C++-style cast:

int *ptoi = reinterpret_cast<int*>(17);

to make it clear what kind of conversion you are trying to invoke

回答4:

The compiler will interpret int *ptoi = 17; as a variable ptoi that points to an integer at location 17 (0x00000011).

C is simply a completely different language. It may look like C++, but it's not. Different rules apply to both languages.

I should probably point out that both C and C++ will (as far as I know) make it a pointer to 0x00000011, but C simply complains less, since assigning memory locations is something that's not invalid in C.

来源：https://stackoverflow.com/questions/8288502/why-can-i-implicitly-convert-an-int-literal-to-an-int-in-c-but-not-in-c

标签

c++

pointers

casting