Thread: Help for understand pointer

Hello,

I'M a newbie C programming,i read book to learn C,in chapter of pointers and arrays, written an example that i write :

Code:

#include <stdio.h>
 
void main()
{
	char test[3][3] = {{4,5,1},{6,9,7},{3,2,8}};
	for(int i = 0;i<9;i++){
			printf("%d\n",*(*test + i));
	}
}

output this code is all of tests elements.

in the line printf("%d\n",*(*test + i)); , in section (*test + i) in section c in each loop i addition to address of first byte of test array.t type is char and each element have 1 byte,thus when address of first byte addition to i,in each loop we go to the next element of array.i changed type of test to int,and compiled this sample again, when type of test is int each element have 4 bytes,thus when *test + i we shouldn't go to next element,but when i printf address of *test + i, i saw the address grows 4 bytes !!
why after i change type of test,when i addition i to *test, address grows 4 bytes ?
i can't understand what happened

excuse me for my bad english .
thanks in advance .

Why are you defining test as an array of 3 arrays of 3 chars when you're accessing it as if it's an array of 9 chars? Your problem is a misunderstanding of types.

in section (*test + i) in section c in each loop i addition to address of first byte of test array.

No, you're accessing i chars from test[0]. Think about the types of each expression.

test + 0: char (*)[3]
test[0] + 0, *(test + 0) + 0: char *
test[0][0], *(*(test + 0) + 0): char

The second one is the one that applies here. You're pointing *into* test[0] which is exactly three chars wide. Dereferencing test[0][3], or anything beyond that, is incorrect.

t type is char and each element have 1 byte,thus when address of first byte addition to i,in each loop we go to the next element of array.

Why do you assume that the first element of the second array is the next element after the last element of the first array? They are separate arrays.

why after i change type of test,when i addition i to *test, address grows 4 bytes ?

Isn't it obvious? sizeof (int) happens to be four bytes on your C implementation, and sizeof (char) is always one. test[0] + 1 adds (sizeof (int)) to test[0] when you define test as an int[3][3]. If you were to do test + 1, it would be scaled by (sizeof (int[3])).

I suggest fixing your code so that it's correct. You could either change the type of test to “T test[9]” and dereference each element with either “*(test + i)” or “test[i]”, or keep the type you have and use two indices, like so:

Code:

int i, j;

for (j = 0; j < 3; j++)
        for (i = 0; i < 3; i++)
                printf("%d\n", test[j][i]);

Otherwise your code is simply incorrect.

Thanks to replies, i can't understand properly pointers and types, what is you're suggestions to learn properly pointers and types ?

I read this :
Beginning C: From Novice to Professional (Beginning Apress): Ivor Horton: 9781430243625: Amazon.com: Books

Originally Posted by Salem

Whilst flattening a 2D array into a 1D array is something that you can do

If you're referring to the way he's accessing the char[3][3], it yields undefined behaviour. Check section 2 of annex J.

An array subscript is out of range, even if an object is apparently accessible with the given subscript (as in the lvalue expression a[1][7] given the declaration int a[4][5])

I suppose it may be feasible to memcpy the data into a char[9], but you can only dereference three pointers from the offset of test[0].

What's your suggestion for c programming book,and a reference for learning types in c ??