Use pointer for multidimension array

[hqt]

I have learnt how to use pointer for multidimension array. this is my example:

int trace [3][4];
int (*ptr)[4] =trace;

After that,I don't know how to use pointer ptr to process elements of trace array.
who can teach me, please.

thanks for tons

[manasij7479]

>int(*ptr)[4] =trace;
This... AFAIK... is useless.
You can already use trace as a pointer.
"trace[i][j]" gives the required element, so does " *(trace+i*4+j) "

[hqt]

Yes, I know this way too. there three ways same same like that to touch trace[i][j]:

1) *(*(trace +i)+j)
2)*(trace[i]+j)
3)*(trace+i)[j]

these ways easy to understand why. but the above example, in my book say that, but they don't say how to use this way
I think will exist a method to use this

@: manasij: I see that you like to use slang to talk today, I have learnt word: asap: as soon as possible. and now, new word:afaik

[MK27]

afaik = as far as I know. You can find most of these by googling "acronym afaik".

I agree with manasij7479 that using a pointer here is generally more awkward than using the index, but there are times when a pointer is easier or tidier. So if you need an example of how to walk a matrix that way:

#include <iostream>

using namespace std;

int main(void) {
	int ray[3][3] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9} },
		(*p1)[3] = ray, *p2 = p1[0];

	while (p1 - ray < 3) {
		if (p2 - *p1 == 3) {
			cout << endl;
			p1++;
			p2 = p1[0];
		} else {
			cout << *p2 << ' ';
			p2++;
		}
	}

	return 0;
}

It is very important when doing this that the pointers are correctly typed!

Some things to notice: p1 is effectively the same type as "ray" (technically, ray is an array while p1 is just a pointer, but ray implicitly converts to a pointer in context). However, p2 is not the same type as p1, and so when doing arithmetic with addresses, you must dereference (via *) p1 to make them equivalent, as on line 10 (just as you must dereference p2 on line 15 to get a value).

[hqt]

Oh, thanks very much to help me solve my problem. I have understand your code. It likes this idea:

int trace[4];
int ptr*=trace; 
// So: (ptr+1)=trace[1]; (ptr+2)=trace[2];

And, I have an other question bulding from above problem:
I have this code:

   int trace[3][4];
   int (*ptr)[3]=trace;

the complier will notice this error:cannot convert 'int (*)[4]' to 'int (*)[3]' in initialization.
I know C++ associate demension of array with type of variable, and int(*)[4] and int(*)[3] cannot convert together. But, why this line:

 int(*ptr)[3]=trace

, C++will know type of trace is int(*)[4]. Can you teach me more, please.

thanks

[laserlight]

An array is converted to a pointer to its first element. So, when you have an array of 3 arrays of 4 ints, it is converted to a pointer to an array of 4 ints, i.e., int (*)[4]. This is different from, and incompatible with, a pointer to an array of 3 ints.