为什么“ int * ptr = int()”值初始化不是非法的？

以下代码(取自从这里) :

int* ptr = int();

在 VisualC + + 中编译并初始化指针。

这怎么可能？我的意思是 int()产生一个类型为 int的对象，我不能将一个 int赋给一个指针。

为什么上面的代码不是非法的？

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The expression int() evaluates to a constant default-initialized integer, which is the value 0. That value is special: it is used to initialize a pointer to the NULL state.

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Because int() yields 0, which is interchangeable with NULL. NULL itself is defined as 0, unlike C's NULL which is (void *) 0.

Note that this would be an error:

int* ptr = int(5);

and this will still work:

int* ptr = int(0);

0 is a special constant value and as such it can be treated as a pointer value. Constant expressions that yield 0, such as 1 - 1 are as well allowed as null-pointer constants.

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Well int isn't an object.

I beleive what's happening here is you're telling the int* to point to some memory address determined by int()

so if int() creates 0, int* will point to memory address 0

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最佳答案

int() is a constant expression with a value of 0, so it's a valid way of producing a null pointer constant. Ultimately, it's just a slightly different way of saying int *ptr = NULL;

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From n3290 (C++03 uses similar text), 4.10 Pointer conversions [conv.ptr] paragraph 1 (the emphasis is mine):

1 A null pointer constant is an integral constant expression (5.19) prvalue of integer type that evaluates to zero or a prvalue of type std::nullptr_t. A null pointer constant can be converted to a pointer type; the result is the null pointer value of that type and is distinguishable from every other value of object pointer or function pointer type. Such a conversion is called a null pointer conversion. [...]

int() is such an integral constant expression prvalue of integer type that evaluates to zero (that's a mouthful!), and thus can be used to initialize a pointer type. As you can see, 0 is not the only integral expression that is special cased.