Comments by Maarten ten Velden (Top 5 by date)

Outputting *a as %d here seems wrong as *a is %p. Doing *a++ is also wrong, I think this would increment a (as postfix ++ operator has higher precedence than * increments a not *a, though *a before increment is still supplied to printf) in steps of sizeof(int*), thus on 64 bit platform (with 32 bit int) would output (? to indicate unknown/undefined value)..

a = 0x?, *a = 1
a = 0x?, *a = 5
a = 0x?, *a = 9
a = 0x?, *a = ?
a = 0x?, *a = ?

You are talking about a multidimensional array though, I am using a non multidimensional array. The problem in your example is that ptr[1] results in a int* and then ptr[1][1] tries to dereference a pointer whereas mat[1][1] calculates an offset right?

My case with a non multidimensional array also causes an error -1073741819, which I assume is STATUS_ACCESS_VIOLATION (thus the segmentation fault you mentioned). Alignment shouldn't be a problem as I am not using a multidimensional array?

I think I do understand now why this doesn't work. I wrote my own solution.

I didn't mean avoiding malloc and free in general, I meant in this particular case for which I am asking the question.

The following is the workaround I meant which compiles..

int** a;
int b[5];
int* c = b;
a = &c;
(*a)[1] = 123;
printf("b[1] == %i\n", (*a)[1]);

The cases you mention seem to be arbitrary rules, as though printf("%u\n", *b) doesn't work, printf("%u\n", b[0]) and int* c = b; printf("%u\n", c); would both achieve the exact same. The last example is entirely different than taking the address of an array, which is a variable, as there you are taking the address of an intermediate result. I still don't see a reason why a C compiler would not be able to directly use the & and * operators on an array.

Well avoiding malloc and free is wanted. You said "justifiably". How is this "justifiably"? I can work around it by assigning b to an int* and then assigning the address of that int* to a, but I don't see why that is workaround is required.