int arr[2][3] = { {1,2,3},{4,5,6} };
int(*parr)[3];
parr = arr;
I'm not sure what int (*parr)[3]; means in the above code.
The arr array declared in the first line consists of two rows and three rows
(*parr)[3] does not have the address value of the first element of each row?
(Address output in hex) - As you can see, parr[0] has the same address as parr[0][0], parr[1] and parr[1][0]. However, parr[2] has an address after arr[1][2] that I didn't generate, and when I checked the value I had at that address, I confirmed that an uninitialized value came out. However, if you declare int (*parr)[2];, a compilation error occurs. I don't understand much about language c, so please explain it in detail It's quite complicated. Thank you!
int (*parr)[3]; is a pointer to a two-dimensional array.
parr = arr; will point to an array of 2*3.
You can access the arr with a pointer as shown below.
[cling]$ parr[0]
(int [3]) { 1, 2, 3 }
[cling]$ parr[1]
(int [3]) { 4, 5, 6 }
And parr[2] goes beyond the range of 2*3 of arr, so you'll get an incorrect value. It's not worth the garbage.and so on
int (*parr)[2]; is a grammatical error in parr = arr;. Since the array is 2*3, 3 spaces are required. So it's a pointer to an array of three elements.
int **pa = (int**)arr You can hold addresses with double pointers, but the array is not accessible. Usually used to point to an array of pointers.
I can't write down the lecture now, but I'm attaching a well-organized address instead.
https://noname2.tistory.com/102
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