C中三元运算符采用不同操作数类型的原理

Question

有很多种类型转换来确保赋值工作，例如隐式类型转换（提升）和显式类型转换（截断），但我不确定它如何在三元运算符的指针类型转换上工作。

#include <stdlib.h>

int main (void)
{
    (void)((rand() ? (char*)NULL :        NULL) + 1);
    /*     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^        -> expression A */

    /* GCC Warning: pointer of type ‘void *’ used in arithmetic */
    (void)((rand() ? (char*)NULL : (void*)NULL) + 1);
    /*     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^        -> expression B */

    return 0;
}

显然，编译器将表达式 A 视为 char* 的类型，而将 B 视为 void* 的类型。

我有两个问题：

我查看了预处理代码，NULL正好展开为((void*)0)，为什么((void*)0)和(void*)((void*)0)是不同的类型?
根据表达式 B，为什么编译器将 char* 的类型转换为 void* 的类型，而不是相反？

Answer 1

我会尝试解释一下：

0 或 (void*)0 是空指针常量，由三元条件运算符适当处理：

if one operand is a null pointer constant, the result has the type of the other operand; (6.5.15 6)

但是(void*)((void *)0)是一个空指针，而不是空指针常量(NULL)：

An integer constant expression with the value 0, or such an expression cast to type void *,is called a null pointer constant. If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function. (6.3.2.3 3)

因此，此段现在适用：

otherwise, one operand is a pointer to void or a qualified version of void, in which case the result type is a pointer to an appropriately qualified version of void. (6.5.15 6)

Answer 2

以下适用：

6.5.15 Conditional operator

The first operand shall have scalar type.
One of the following shall hold for the second and third operands:

/--/

— both operands are pointers to qualified or unqualified versions of compatible types;
— one operand is a pointer and the other is a null pointer constant; or
— one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void.

在第一种情况下，您有一个操作数 (char*)NULL，它是一个（非限定的）类型指针，还有一个操作数 NULL，它是一个空指针常量。

在 §6 中更进一步说：

If both the second and third operands are pointers or one is a null pointer constant and the other is a pointer, the result type is a pointer to a type qualified with all the type qualifiers of the types referenced by both operands

用简单的英语来说：如果一个操作数是 char* 类型，另一个是空指针常量，则结果是 char*.

在第二种情况下，您有一个指向类型的指针和一个 void*。空指针常量也不是（我将在下面进一步解释原因）。在同一段落的下方，它说（强调我的）：

... if one operand is a null pointer constant, the result has the type of the other operand; otherwise, one operand is a pointer to void or a qualified version of void, in which case the result type is a pointer to an appropriately qualified version of void.

表示第二个?:操作的结果是void*类型。这应该回答你的第二个问题。

要回答你的第一个问题，它与 ?: 运算符本身无关，而是关于空指针与空指针常量的 "ancient C mystery"。

如何扩展 NULL 宏是实现定义的，它是 0 或 (void*)。它保证是一个空指针常量。

，即空指针可以具有任何实现定义的值，而空指针常量始终是 0 或 (void*)0.

6.3.2.2 说：

An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant. If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function.

所以(void*)NULL是空指针，但不是空指针常量。因此，条件运算符将其视为空指针。

总结一下：空指针不是条件运算符的特例，只有空指针常量才是。

C中三元运算符采用不同操作数类型的原理

Principle of different operand types adoption for ternary operator in C

c

pointers

casting

null

ternary-operator