First off, the primer on multi dimensional arrays: https://www.tutorialspoint.com/cplusplus/cpp_multi_dimensional_arrays.htm
This tutorial on array pointers seems lame at first because it starts with just pointers to 1D arrays which wasn't helping me, then I realized that I have to scroll down farther. Note: Check out the sizeof examples somewhere near the end: https://www.geeksforgeeks.org/pointer-array-array-pointer/
This link covers the exact same material, but somehow it seems less opaque. It was from this that I realized that what I needed was to define a pointer as a 1d array pointer of the same number of columns as the 2d array to get what I wanted. And then somehow it works to use that pointer with 2 indexes.https://www.programiz.com/cpp-programming/pointers-arrays
The wags at StackOverflow helped me learn the trick to finding the size of an array: https://stackoverflow.com/questions/10274162/how-to-find-2d-array-size-in-cHere is the demo program that I tested out on the mbed simulator that has all of the erroneous ways of using array pointers removed, thanks to the above links:
int testme[3][4] = { /* rows, columns */Result of the above:
{0, 1, 2, 3} , /* initializers for row indexed by 0 */
{4, 5, 6, 7} , /* initializers for row indexed by 1 */
{8, 9, 10, 11}
};
int main() {
int arr[5] = { 1, 2, 3, 4, 5 };
int *ptr = arr;
int *ptr2;
int (*ptr3)[4];
ptr2 = &testme[0][0];
ptr3 = testme;
printf("%p\n", ptr);
printf("%d\n", ptr[1]);
printf("%d\n", *(*(testme + 2)));
printf("%d\n", *(ptr2 + 5));
printf("%d\n", *(ptr2 + 1));
printf("%d\n", ptr3[1][2]);
printf("array number of rows: %d\n",sizeof(testme)/sizeof(testme[0]));
printf("array number of columns: %d\n",sizeof(testme[0])/sizeof(int));
return 0;
}
0x2e60
2
8
5
1
6
array number of rows: 3
array number of columns: 4
