C/C++:指针算术

C/C++: Pointer Arithmetic(C/C++:指针算术)

问题描述

我在 Pointer Arithmetic 中读了一点，我发现了两件事我不明白，也不知道它的用途

I was reading a bit in Pointer Arithmetic, and I came upon 2 things I couldn't understand neither know it's use

address_expression - address_expression

还有

address_expression > address_expression

谁能给我解释一下，它们是如何工作的以及何时使用.

Can someone please explain them to me, how do they work and when they are used.

我想说的是如果我只取两个地址并减去它们会产生什么

What I meant to say is what do they produce if I just take two addresses and subtract them

如果我取两个地址并比较它们是什么结果或基于比较

And If I take two addresses and compare them what is the result or comparing based upon

减去地址的结果我现在明白了，但是比较地址还是不明白.

I now understand the result of subtracting addresses, but comparing addresses I still don't get it.

我知道 1<2，但是一个地址如何大于另一个地址以及它们的比较对象是什么

I understand that 1<2, but how is an address greater than another one and what are they compared upon

推荐答案

指针减法产生相同类型的两个指针之间的数组元素数.

Pointer subtraction yields the number of array elements between two pointers of the same type.

例如

int buf[10] = /* initializer here */;

&buf[10] - &buf[0];  // yields 10, the difference is 10 elements

指针比较.例如，对于 > 关系运算符:如果左侧的指向数组元素或结构成员，则 > 操作产生 1在右侧的指向数组元素或结构成员之后，否则产生 0.记住数组和结构是有序序列.

Pointer comparison. For example, for the > relational operator: the > operation yields 1 if the pointed array element or structure member on the left hand side is after the pointed array element or structure member on the right hand side and it yields 0 otherwise. Remember arrays and structures are ordered sequences.

 &buf[10] > &buf[0];  // 1, &buf[10] element is after &buf[0] element

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