Multi-dimensional array and pointer

When I wrote 3.43, I thought about the answer for two days. I would like to record it.

 

The topic is to compile three different versions of programs so that they can output Ia elements. Version 1 uses the range for statement, version 2 uses the common for + subscript operator, and Version 3 uses the for statement + pointer.

Below are the results of the two-day trial...

#include<iostream>using namespace std;int main(){    int ia[2][3] = {1,2,3,3,2,1};    //version 1    for(int (&i1)[3]:ia){        for(int j1:i1)            cout<<j1<<" ";    };    cout<<endl;         //version 2    for(int i2 = 0;i2 < 2;++i2){        for(int j2 = 0;j2 < 3;++j2)            cout<<ia[i2][j2]<<" ";    }    cout<<endl;        //version 3    for(int (*i3)[3] = ia;i3 != &ia[2];++i3){ //also  for(int (*i3)[3] = begin(ia);i3 != end(ia);++i3){        for(int *j3 = *i3; j3 != *i3+3 ;++j3) //also  for(int *j3 = begin(*i3); j3 != end(j3);++j3)        　　cout<<*j3<<" ";    }        return 0;}

 

If version 2 is relatively easy to write, I will not repeat it here. It is mainly about version 1 and version 3.

Version 1:

At the beginning, it seems that I directly use an int * pointer to traverse IA, and then I printed only 1 and 3 at the end of the flower line. I did not know it until I wrote it to version 3 ), in the multi-dimensional array {1, 2, 3}, {3, 2, 1}, Ia points to the first element, while IA + 1 points, points to the first element of the second one-dimensional array. Therefore, only one pointer is obviously unable to print all elements. Return to the textbook. There is an example, INT (* P) [4] = IA, P points to an array containing four integers (IA [3] [4] in the textbook). So I was wondering if, once defined in this way, a [4] is added, it will open up a space of four integers for P. Then I added [3] to the initial version of version 1 and a dual loop. Obviously, the second range is a pointer and compilation error... Then, according to the book, I changed * To &, so I finally did not skip error...

PS: there is another problem. In the for loop, the item type can be various, which can be a pointer or Int. Since the book always uses auto to let the compiler recognize it, I mistakenly thought that item can only be a pointer... Actually, it can be a variety of things.

 

 

Version 3:

Version 3 is really tangled. At first I wanted to save time using begin (IA) and end (IA), and I didn't know the problems of IA and IA + 1 at first, the results show that the types in the brackets do not match. Therefore, the begin and end methods are abandoned and the IA and IA + n methods are used. Then I learned about the problem mentioned in the red letter above, so I immediately changed a loop, so OK, a loop no longer jumps error. So the double loop is written as int * J3 = I3 at the beginning. I think it's all pointers. So there should be no problem in writing, so there is another problem... So I continued to change it. I tried to add and add... As a result, I checked the data line again. Mom's egg processes each one-dimensional array, and then adds * to process every element in the one-dimensional array. OK, so the problem is solved. Just now I want to add begin and end, and I find that there is no problem at all. After knowing the two problems of the red letter, I will solve them all at once. It's easy to use begin and end, saving a lot of effort on your fingers!

 

So after the competition, I understood a question. I should study more materials and don't always look for answers... I don't know what people think when I find the answer...