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?

编程入门行业动态更新时间:2024-10-28 06:21:47

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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.