C ++中的指针递增 [英] Pointer incrementing in C++

问题描述

这是什么意思:指针增量指向该指针的下一个基本类型的地址?
例如:

p1++;  // p1 is a pointer to an int

此语句是否意味着p1指向的地址应更改为下一个int的地址，或者仅应将其递增2(假设int为2个字节)，在这种情况下，特定地址可能不包含int?
我的意思是，如果p1是0x442012，则p1++将是0x442014(可能是双精度地址的一部分)，还是会指向下一个int，该地址位于类似0x44201F的地址中? /p>

谢谢

解决方案

指针算法并不关心指针对象的内容或有效性.它将使用以下公式简单地增加指针地址:

new_value = reinterpret_cast<char*>(p) + sizeof(*p);

(假定指向非const的指针–否则强制转换将不起作用.)

也就是说，无论指针值和内存对齐如何，它都会使指针增加sizeof(*p)个字节.

What does this mean: that a pointer increment points to the address of the next base type of the pointer?
For example:

p1++;  // p1 is a pointer to an int

Does this statement mean that the address pointed to by p1 should change to the address of the next int or it should just be incremented by 2 (assuming an int is 2 bytes), in which case the particular address may not contain an int?
I mean, if p1 is, say, 0x442012, will p1++ be 0x442014 (which may be part of the address of a double) or will it point to the next int which is in an address like 0x44201F?

Thanks

解决方案

Pointer arithmetic doesn’t care about the content – or validity – of the pointee. It will simply increment the pointer address using the following formula:

new_value = reinterpret_cast<char*>(p) + sizeof(*p);

(Assuming a pointer to non-const – otherwise the cast wouldn’t work.)

That is, it will increment the pointer by an amount of sizeof(*p) bytes, regardless of things like pointee value and memory alignment.

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