为什么我能隐式转换为int的文字为int *在C，但不是在C ++？ [英] Why can I implicitly convert an int literal to an int * in C but not in C++?

问题描述

我认为在以下code，C自动转换17到为int * ，正如最近有人指出，（但没有给原因为何），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，如果我做如上C ++同样的事情，我从int到INT * 得到一个错误说无效的转换。但是，如果我在做C ++以下，正常工作

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;

这是我认为，在C，铸造是隐含的原因。

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

有人能解释为什么，在C ++中，我一定要投它但在C，它工作正常？

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

推荐答案

转换，而不强制转换也是C.非法大多数编译器将让您摆脱它虽然。铛发出警告：

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说，在栏目6.5.4转换操作符第4段：

这涉及到指针，不包括依据的6.5.16.1限制允许的其他转换，由显式类型转换的方式来指定。

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是例外无效* 转换为其它指针而不需要转换。

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

C ++的规范说，在栏目5.4显式类型转换（转换符号），第3段：

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

下面没有提及，而不是由用户明确定义的任何类型的转换是非法的构造。

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

所以你去 - 在两种语言中是非法的，但有很多旧的软件兼容性，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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