Passing a multi-dimensional array to a function in C

Question

I wrote this function to swap values in a multi-dimensional array with my understanding that arrays are pointers.

void
swap(int* a, int* b)
{
    int tmp = *a;
    *a = *b;
    *b = tmp;
}

However when I try to use the function

swap(board[d-1][d-2]), board[d-1][d-33];

I get these errors from the compiler and I don't know why:

fifteen.c: in function 'init':
fifteen.c:166:9: error: passing argument 1 of 'swap' makes pointer from integer without a cast [-werror]
fifteen.c:45:6: note: expected 'int *' but argument is of type 'int'
fifteen.c:166:9: error: passing argument 2 of 'swap' makes pointer from integer without a cast [-werror]
fifteen.c:45:6: note: expected 'int *' but argument is of type 'int'

How do I fix it?

@chris what are they? mmm IMHO, I thought the where a big chuck of memory that you navigate through according to the data type being handles, ex.: int64_t num = 1234; and memcpy(buf, &num, sizeof(num)) in this case num behaves as an array of 8 bytes. ex.: char text[] = "..."; then int32_t *ptr=(int32_t*)&text[0]; and when you ptr++; the content of *ptr will be the next 4 bytes inside text. — yeyo
– yeyo, Commented Oct 27, 2012 at 18:36
@Kira, Arrays can decay into pointers, but they are not pointers themselves. Try comparing sizeofs: liveworkspace.org/code/853fc936ff59eee2994ab28ca0c02e92 — Qaz
– Qaz, Commented Oct 27, 2012 at 18:43
@chris, thanks for the example. It worked as you claim, but I won't lie, I'm a little skeptical about this right now. You're making me doubt XD, now I feel like I need to make an open question about this to convince myself. Thank you very much for pointing this out. — yeyo
– yeyo, Commented Oct 27, 2012 at 19:02
@Kira, There are some good array resources on this site and on the web. — Qaz
– Qaz, Commented Oct 27, 2012 at 19:05

md5 · Accepted Answer · 2012-10-27 17:13:40Z

3

board[d-1][d-2] and board[d-1][d-33] are int. To swap the both, you have to pass their addresses:

swap (&board[d - 1][d - 2]), &board[d - 1][d - 33];

If you are using swap (board[d - 1][d - 2]), &board[d - 1][d - 33]), the instruction int tmp = *a; will try to access to the value on the address board[d - 1][d - 2]: this make no sense! Because you are using pointers, you have to pass the address of your variables.

edited Oct 27, 2012 at 17:13

answered Oct 27, 2012 at 17:06

md5

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alk · Accepted Answer · 2012-10-27 17:06:32Z

2

You need to pass the addresses (swap() expects two int *):

swap (&board[d-1][d-2], &board[d-1][d-33]);

answered Oct 27, 2012 at 17:06

alk

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dead programmer · Accepted Answer · 2012-10-27 17:55:34Z

0

you are passing wrong argument

swap (board[d-1][d-2]), board[d-1][d-33]);

you have to pass the address of the variable actually.So the correct way of calling this function is below.

swap (&board[d-1][d-2]), &board[d-1][d-33]);

I think this will help you.

answered Oct 27, 2012 at 17:55

dead programmer

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