是if（pointerVar）和if（pointerVar！= NULL）一样吗？ [英] Is if(pointerVar) the same as if(pointerVar!=NULL)?

问题描述

简单问题：

是 if（pointerVar）与 if pointerVar！= NULL）？

此外， if（！pointerVar）与相同if（pointerVar == NULL）？

给我最技术上正确/这两个语句似乎和有意义的操作相同。

解决方案

对于最迂回的答案，这里是相关的

首先，如果语句工作，从& p>

作为switch语句之外的语句中初始化声明的条件的值是声明变量的值隐式转换键入 bool 。

但是指针如何转换为 bool s？你可以问。好的，这里是&; 4.12.1：： - ）

算术，枚举，指针或指向成员类型的指针可以转换为bool类型的右值。零值空指针值或空成员指针值将转换为false;任何其他值都将转换为true。

这意味着语句

  if（ptr）
  

>

  if（（bool）ptr）
  

这等价于

  if（ptr == NULL）
  
 
 
 
但
 如果（！ptr）
  
好吧，C ++ spec和5.3.1.8说， p> 
 
 
 
 
逻辑否定运算符的操作数！隐式转换为 bool （第4条）;如果转换的操作数 false 和 false  true  >否则。结果的类型是 bool 。
 
 
 p> 
 
 
  if（！ptr）
  
 
 b $ b 
等同于
  if（！（bool）ptr）
  
 
 
 
这等同于
  if（！（ptr == NULL））
  
这最终相当于
  if（ptr！= NULL）
  
 
 $ b b 
哇！这是一个有趣的搜索。希望这回答你的问题！
 
 
 
当然，这个故事还有更多。  NULL 不是C ++语言的一部分;它是< cstddef> 中的一个宏定义为
  #define NULL 0 
  
这样做是因为C ++标准在§ 4.10.1中定义了空指针
 
 
空指针常数是整数类型的整数常数表达式（5.19），其值为零。空指针常量可以转换为指针类型;结果是该类型的空指针值，并且可以与指向对象的指针或指向函数类型的指针的每个其他值区分开。
 
 
更正确，我应该使用数字字面值0在上面的例子。但是，如果你包含< cstddef> ，那么在预处理之后，这个代码就会生成相同的代码。
 
Simple question:

Is if (pointerVar) the same as if (pointerVar!=NULL)?

Also, is if (!pointerVar) the same as if (pointerVar==NULL)?

Give me your most technically correct/pedantic answer. The two statements seem and make sense to operate the same. Is there anything wrong with the former though (besides its slightly lower readability)?
解决方案
For the most pedantic answer, here's the relevant sections of the spec.

First, here's how if statements work, from §6.4.4:

  
The value of a condition that is an initialized declaration in a statement other than a switch statement is the value of the declared variable implicitly converted to type bool. If that conversion is ill-formed, the program is ill-formed.
"But how are pointers converted to bools?" you may ask.  Well, here's §4.12.1: :-)

  
An rvalue of arithmetic, enumeration, pointer, or pointer to member type can be converted to an rvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true.
So this means that the statement
if (ptr)
is equivalent to
if ((bool) ptr)
which is in turn equivalent to
if (ptr == NULL)
But what about
if (!ptr)
Well, the C++ spec, §5.3.1.8, says that

  
The operand of the logical negation operator ! is implicitly converted to bool (clause 4); its value is true if the converted operand is false and false otherwise. The type of the result is bool.
So this means that
if (!ptr)
is equivalent to
if (!(bool)ptr)
which is in turn equivalent to
if (!(ptr == NULL))
which is finally equivalent to
if (ptr != NULL)
Whew!  That was a fun search to do.  Hope this answers your question!

Of course, there is more to this story.  NULL is not part of the C++ language; it's a macro in <cstddef> defined as
#define NULL 0
This works because the C++ standard defines the null pointer in §4.10.1 as

  
A null pointer constant is an integral constant expression (5.19) rvalue of integer type that evaluates to zero. 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 pointer to object or pointer to function type
So to be more correct, I should have been using the numeric literal 0 in the above examples.  However, if you have <cstddef> included, then this works out to the same code after preprocessing.

                        
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