Accessing elements in a static array using pointer(arithmetic) in C

Question

If I have the following code in a function:

int A[5][5];
  int i; int j;
  for(i=0;i<5;i++){
    for(j=0;j<5;j++){
      A[i][j]=i+j;
      printf("%d\n", A[i][j]);
    }
  }

This simply prints out the sum of each index. What I want to know is if it's possible to access each index in the static array in a similar fashion to dynamic array. So for example, if I wanted to access A[2][2], can I say:

*(A+(2*5+2)*sizeof(int))?

I want to perform some matrix operations on statically allocated matrices and I feel like the method used to dereference dynamic matrices would work the best for my purposes. Any ideas? Thank you.

The multiplication by sizeof() is unnecessary if the pointer being dereferenced is a pointer-to-int. Pointer arithmetic steps by the size of the type pointed to pointed. — dmckee --- ex-moderator kitten
– dmckee --- ex-moderator kitten, Commented Feb 28, 2012 at 20:48
@dmckee not only is it not necessary, it's wrong. First, it won't do what he's expecting it does, second, you can go over the bound of the array and get into undefined behavior. — Luchian Grigore
– Luchian Grigore, Commented Feb 28, 2012 at 20:50
To @EdS.'s point the array is "automatic" or "at file scope" depending on the existence or non-existence of a un-exhibited enclosing function. — dmckee --- ex-moderator kitten
– dmckee --- ex-moderator kitten, Commented Feb 28, 2012 at 20:50

Luchian Grigore · Accepted Answer · 2012-02-28 20:45:45Z

4

That's the way to do it: A[i][j].

It prints out the sum of the indexes because, well, you set the element A[i][j] to the sum of the indexes: A[i][j] = i+j.

answered Feb 28, 2012 at 20:45

Luchian Grigore

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3 Comments

Adam Liss Over a year ago

I think the OP is asking how to access an element using pointer arithmetic. It's more a question about the way the data is allocated than about the "proper" way to access it.

Luchian Grigore Over a year ago

@AdamLiss could be, or maybe he's confused that he's getting the sum of the indexes.

ruakh Over a year ago

@LuchianGrigore: I read the statement, "This simply prints out the sum of each index" a bit differently from how you read it. You read it as, "The problem is, this is supposed to print out the array elements, but instead, it prints out the sums of their indices." I read it as, "This is just simple code to show what I mean; it prints out the array elements, which are set to the sums of their indices."

ouah · Accepted Answer · 2012-02-28 20:46:05Z

1

You can use:

*(*(A + 2) + 2)

for A[2][2]. Pointer arithmetics is done in unit of the pointed type not in unit of char.

Of course, the preferred way is to use A[2][2] in your program.

answered Feb 28, 2012 at 20:46

ouah

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2 Comments

ouah Over a year ago

@LuchianGrigore I assumed the OP asked the equivalence of A[2][2] with explicit pointer arithmetic.

Luiggi Mendoza Over a year ago

@LuchianGrigore A[2][2] its not wrong. But he's asking a way to access the elements of A using its pointer value.

John Bode · Accepted Answer · 2012-02-28 21:33:35Z

The subscript operation a[i] is defined as *(a + i) - you compute an offset of i elements (not bytes) from a and then dereference the result. For a 2D array, you just apply that definition recursively:

a[i][j] == *(a[i] + j) == *(*(a + i) + j)

If the array is allocated contiguously, you could also just write *(a + i * rows + j).

When doing pointer arithmetic, the size of the base type is taken into account. Given a pointer

T *p;

the expression p + 1 will evaluate to the address of the next object of type T, which is sizeof T bytes after p.

Note that using pointer arithmetic may not be any faster than using the subscript operator (code up both versions and run them through a profiler to be sure). It will definitely be less readable.

user995502 · Accepted Answer · 2012-02-28 23:35:36Z

Pointer arithmetic can be tricky. You are on the right track, however there are some differences between pointer and normal arithmetic. For example consider this code

int I = 0;
float F = 0;
double D = 0;
int* PI = 0;
float* PF = 0;
double* PD = 0;

cout<<I<<" "<<F<<" "<<D<<" "<<PI<<" "<<PF<<" "<<PD<<endl;
I++;F++;D++;PI++;PF++,PD++;

cout<<I<<" "<<F<<" "<<D<<" "<<PI<<" "<<PF<<" "<<PD<<endl;
cout<<I<<" "<<F<<" "<<D<<" "<<(int)PI<<" "<<(int)PF<<" "<<(int)PD<<endl;

If you run it see the output you would see would look something like this (depending on your architecture and compiler)

0 0 0 0 0 0
1 1 1 0x4 0x4 0x8
1 1 1 4 4 8

As you can see the pointer arithmetic is handled depending on the type of the variable it points to.

So keep in mind which type of variable you are accessing when working with pointer arithmetic.

Just for the sake of example consider this code too:

void* V = 0;

int* IV = (int*)V;
float* FV = (float*)V;
double* DV = (double*)V;

IV++;FV++;DV++;
cout<<IV<<" "<<FV<<" "<<DV<<endl;

You will get the output (again depending on your architecture and compiler)

0x4 0x4 0x8

Remember that the code snippets above are just for demonstration purposes. There are a lot of things NOT to use from here.

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