问题

我对指针的%p和%v格式化动词提供的输出感到困惑。

var a *int
// a的零值为：
fmt.Printf("%v", a) // <nil>
// 使用%p
fmt.Printf("%p", a) // 0x0

我们知道%v默认使用%p作为格式化器，那么为什么%v的输出是<nil>呢？

未初始化指针的零值是<nil>，那么为什么%p输出的是0x0而不是<nil>呢？

根据Go文档，%p使用带有前导0x的16进制表示法，那么<nil>和0x0是相同的吗？

英文:

I am confusing the outputs provided by %p and %v format verbs of a pointer

var a *int
// zero value of a is:
fmt.Printf(&quot;%v&quot;, a) // &lt;nil&gt;
// using %p 
fmt.Printf(&quot;%p&quot;, a) // 0x0

We know that %v uses %p as the default formatter, then why %v showing <nil> as output?

The zero value of an uninitialized pointer is <nil>, then why %p outputting 0x0 instead <nil>?

As per Go docs, %p uses base 16 notation with leading 0x, if so, is <nil> and 0x0 are same?

答案1

得分: 4

（提示：试图通过聪明地使用fmt.Printf来学习东西是非常困难的，因为它包含了很多（合理的）魔法。）

英文:

v is the generic value verb and it prints nil pointers as <nil> because that makes most sense as a generic formatting. Note the word "default" in its description: It uses p as default, not unconditionally.
p prints pointer values and thus prints a nil pointer as 0x0. Note that p and v are different verbs.
Yes, nil pointers are 0. But note that the strings <nil> and 0x0 are not "the same".

(Tip: It is incredibly hard to learn things by trying to be clever with fmt.Printf as it contains a lot of (sensible) magic.)

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