please help array question

[Hoser83]

Hey folks. I'm a little confused here. I'm trying to write some code and I was wondering... if I have a two dimensional array declaration as such:

int A [2][5];

I'm asked to use A[1] in the code. I'm confused because I'm not sure what the type of A[1] is. Is it some kind of pointer to int or is it incorrect syntax and it has no type. Wouldn't A[0][1] be more correct of A[1][0]? Any info would be great.

Thanks!

[Hoser83]

oops. please excuse my mistake in not code tagging that, and I meant

wouldn't A [1][0] or A [0][1] be more correct in accessing an element?

[Salem]

> I'm asked to use A[1] in the code
Like
int *pointerToSecondRow = A[1];

[itsme86]

It depends on what you're actually supposed to do:

This is perfectly valid code:

/* onedim.c */

#include <stdio.h>

void show_one_dim(int *a, int nelems)
{
  int i;

  for(i = 0;i < nelems;++i)
     printf("%d ", a[i]);
  putchar('\n');
}

int main(void)
{
  int a[2][5] = { { 0, 1, 2, 3, 4 }, { 5, 6, 7, 8, 9 } };

  show_one_dim(a[0], sizeof(a[0]) / sizeof(*a[0]));
  show_one_dim(a[1], sizeof(a[1]) / sizeof(*a[1]));

  return 0;
}

My output:

itsme@itsme:~/C$ ./onedim
0 1 2 3 4
5 6 7 8 9
itsme@itsme:~/C$

[Hoser83]

To get me started they just give me:

int A [2][5];

Then it says to use the type A [1] in the code (if the expression has a type, if it is not, state why it's not) . If it does, I have to state what type of expression it is. They ask if it is an int, int*, or a int**. And that's all they give me so I wasn't sure what to do. If you guys could steer me in the right direction, I would really appreciate it.

Thanks

[itsme86]

If you have int A[2][5] then A[1] is type int *.

[Hoser83]

I'm pretty new to C and just so I understand the concept...... is
A [1] a pointer to int all of the time or just in your example of
code? Does A [1] mean that it points to only one dimension of the array? does A [0] point somewhere else? Does
&A[1] mean something else? I just want to be sure that I understand. Thanks for the help man, I really appreciate this.

[Dave_Sinkula]

Given this:

int A [2][5];

Then A[1] has type int [5]. When used in value context it becomes an int *, a pointer to the first element of the array.

An interesting effect then occurs with sizeof, which uses object context.

printf("sizeof A[1] = %d\n", (int)sizeof A[1]); /* sizeof A[1] = 20 */

Arrays are not pointers so if you had this.

int *a = A[1]; /* &A[1][0] */

Then you could not reassign A[1].

A[1] = a; /* error */

(I hope I am getting this right.)

[edit]http://67.40.109.61/torek/c/pa.html