Question

I have a basic doubt in 2D arrays (C Language). Consider a declaration of a 2D array as follows

int array[3][5];

Now when I do the following, the output of both th below printf’s is the same:

printf("%u\n", array);
printf("%u\n", *(array));

Now when try to to the following:

printf("%u\n", array+1);
printf("%u\n", *(array)+1);

The outputs are different. I get that the 2nd printf refers to array[0][1] and the first one to array[1][0]. How does this work? array is a pointer to what?

Thanks in advance

Answer 1

Arrays are not pointers. Ignore any answer, book, or tutorial that tries to tell you otherwise.

An expression of array type, in most contexts, is converted (at compile time) into a pointer to the array’s first element. The exceptions are:

• The operand of sizeof (sizeof arr yields the size of the array, not the size of a pointer)
• The operand of unary & (&arr yields the address of the array, not of its first element — same memory location, different type). This is particularly relevant to your example.
• A string literal in an initializer used to initialize an array object (char s[6] = "hello"; doesn’t copy the address of the string literal, it copies its value)

A 2-dimensional array is nothing more or less than an array of arrays. There are other data structures that can be used with the same x[y][z] syntax, but they’re not true 2-dimensional arrays. Yours is.

The [] indexing operator is defined in terms of pointer arithmetic. x[y] means *(x+y).

The behavior of your code follows from these rules.

Read section 6 of the comp.lang.c FAQ. It’s the best explanation of this stuff I’ve seen.

And don’t use "%u" to print pointer values; convert to void* and use "%p".

printf("%p\n", (void*)array);
printf("%p\n", (void*)*(array));