不明白为什么两个表达式是相同的 [英] Don't understand why two expressions are the same

问题描述

char multi [3] [10];



printf（地址是％p / n，多+ 3）;

printf（地址是％p \ n，*（多+ 3））;



解除引用的地址与地址相同... 。为什么？



*（多+ 3）是否作为数组的名称而是指针？



我尝试了什么：



让程序得到相同的地址......并且不明白怎么样地址

当解除引用时与地址本身相同

char multi[3][10];

printf("Address is %p/n",multi + 3);
printf("Address is %p\n",*(multi + 3));

The dereferenced address is the same as the address .... Why?

Is *(multi + 3) acting as the name of an array and is therefor a pointer?

What I have tried:

Ran the program got the same address ... and dont't understand how an address
when dereferenced is the same as the address itself

推荐答案

这很复杂，但它与什么类型<$ c有关$ c>多是。

是 char ？不 - 显然不是。

是指向 char 的指针？没有。

那么它是什么？

multi 是一个包含三个元素的数组，每个元素都是一个数组10 char 值。

那么多+ 3到底是什么？基本上，这是一个非法地址， 恰好在有效的内存空间 ，因为 multi 只有三个元素： 多 + 0，多 + 1，多 + 2.

因为你根本没有初始化你的数组，所以 multi + 3的值最有可能是第一个 char multi [0] 数组中的实际 char 数据的值c $ c> s，这可能是 - 但不是肯定的，这取决于你的编译器和编译器选项 - 0.



基本上，不要这样做。 br />
尝试使用有效地址，您会得到不同的结果，如您所料。

This is complicated, but it has to do with what type multi is.
Is it a char? No - obviously not.
is it a pointer to a char? No.
So what is it?
multi is an array of three elements, each of which is an array of 10 char values.
So what the heck is "multi + 3"? basically, it's an illegal address, which happens to be within the valid memory space because multi has only three elements: multi + 0, multi + 1, and multi + 2.
And because you don't initialise your array at all, the value of "multi + 3" is most likely to be the first value of the actual char data in the multi[0] array of chars, which is probably - but not definately, it depends on your compiler and compiler options - 0.

Basically, don't do it.
Try it with valid addresses, and you get different results, as you would expect.

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