C中的双指针
有人请详细说明这里发生了什么吗？
int main()
{
int **p = 0;
//p=? and why| *p=? and why|**p=? and why
++p;
//p=? and why| *p=? and why|**p=? and why
printf("%d\n", p);
return 1;
}
输出：-
- 4（为什么？）
Anyone please elaborate what is happining here?
int main()
{
int **p = 0;
//p=? and why| *p=? and why|**p=? and why
++p;
//p=? and why| *p=? and why|**p=? and why
printf("%d\n", p);
return 1;
}
output:-
- 4 (why?)
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评论（5）
首先，
p是一个指向整数的指针。int **p = 0;p= 0，*p= 无，**p= 小于没有什么。++p;与 p = p + 1 相同。进一步表示指向 int 指针的一个指针的大小。至少在您的操作系统上，指针基本上是 32 位长度（4 个字节）。
p现在指向 0 之后的 4 个字节。p的值为 4。First of all,
pis a pointer to a pointer-to-integer.int **p = 0;p= 0,*p= nothing,**p= less than nothing.++p;Same as p = p + 1. Means the size of one pointer to a pointer-to-int further. A pointer is basically, at least on your OS, 32 bits length (4 bytes).
pnow points 4 bytes after 0. The value ofpis 4.p是一个指向int指针的指针。它被初始化为0，即它是一个空指针。然后它会递增以指向内存中下一个连续的指向
int的指针。* 下一个指针将位于地址 4，因为在您的平台上，a 的大小指针为4字节。然后 printf 将指针值解释为整数，因此显示“4”。
* 但请注意，这现在是未定义的行为。
pis a pointer to a pointer-to-int. It's being initialised to0, i.e. it's a null pointer.It's then being incremented to point at the next consecutive pointer-to-
intin memory.* The next pointer will be at address 4, because on your platform the size of a pointer is 4 bytes.Then printf interprets the pointer value as an integer, and so displays "4".
* Note, however, that this is now undefined behaviour.
很明显。您有一个指向 int 的指针（
int **p表示指向 int 的指针），它实际上保存地址 0）。在您的体系结构中，指针本身的长度为 32 位（4 字节），因此递增p会得到 p+4，即 0+4 = 4。去读一本不错的 C 书，然后了解指针算术。你会幸福一辈子的！ :)
It is clear. You have a pointer to a pointer to int (
int **pmeans a pointer to a pointer to int), that actually holds the address 0). A pointer in itself, in your architecture, is 32 bits (4 bytes) long, so incrementingpgives you p+4, that is, 0+4 = 4.Go get a nice C book and learn about pointer arithmetic. You'll be glad the rest of your life! :)
++p实际上是未定义的行为，但在您的实现中似乎发生的是sizeof(int*)为 4，并且空指针是地址 0。回想一下该指针增量（当它不是 UB 时）会向地址添加等于引用对象类型大小的字节数。因此，当您采用int**类型的空指针（因此引用类型为int*）并递增它时，您最终会得到地址4。只是无法保证。当
%d格式需要int时传递指针也是未定义的行为，但似乎int和int 的表示**足够兼容，并且您的实现上的 varargs 调用约定对它们的处理非常相似，因此它已成功打印4。对于 sizeof(int) == sizeof(int**) 的实现来说，这也并不奇怪，但也不能保证。当然，由于这是未定义的行为，因此您所看到的还有其他可能的解释。
++pis actually undefined behaviour, but what appears to have happened on your implementation is thatsizeof(int*)is 4, and a null pointer is address 0. Recall that pointer increment, when it's not UB, adds a number of bytes to the address equal to the size of the referand type. So it's not all that surprising that when you take a null pointer of typeint**(hence the referand type isint*) and increment it, you end up at the address4. It's just not guaranteed.Passing a pointer when the
%dformat expects anintis also undefined behavior, but it appears that the representation ofintandint**are sufficiently compatible, and the varargs calling convention on your implementation treats them sufficiently similarly, that it has successfully printed4. That's also not very surprising for implementations wheresizeof(int) == sizeof(int**), but also isn't guaranteed.Of course since it's undefined behavior, there are other possible explanations for what you see.
p 是一个指向 int 的指针。并且它被初始化为0，即NULL。
当您递增它时，它现在指向指向 int 的下一个指针，在 32 位系统上，该指针恰好是 4。
p is a pointer to pointer to int. And it's initialized to 0, i.e. NULL.
When you increment it, it now points to next pointer to int, which, on 32-bit systems, happens to be 4.