access to an array of pointers in c

Question

Why can I access an array of pointers with two parameters, when it's defined as one-dimensional?

I know, I have to work with an array of pointers to access a multi-dimensional array in a function, but I don't know why I can access the array of pointers with two parameters.

int a[m][l] { 1,2,3,4, 2,3,4,5, 3,4,5,6  }; //some matrices
int c[m][l];    
int *ap[m];  //array of pointers one-dimensional
int i,j;


for (i = 0; i < m; i++)  //passing the address of first element in each 
        ap[i] = a[i];    //colon of matrix to array of pointers

for (j = 0; j < m; j++)
        bp[i] = b[i];

dosomethingwithmatrix(ap[0], bp[0], m, l);



int* dosomethingwithmatrix(const int*ap[], int* bp[])
{
            cp[i][j] = ap[i][j] //accss the array of pointers with two parameters

}

John Hascall · Accepted Answer · 2016-01-15 14:52:38Z

Also in C an array is said to decay into a pointer.

From the standard (C99 6.3.2.1/3 - Other operands - Lvalues, arrays, and function designators):

Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue.

So:

array[i] "decays" to pointer[i]
where pointer has the address of the [0]th element of array

And since we have already seen that:

p[i] == *(p + i)

all we are doing is adding offsets to pointers.

BTW, since addition is commutative, *(p + i) == *(i + p), it gives the sometimes surprising result that:

3["hello world"]

is a perfectly valid C expression (and it equals "hello world"[3]).

Sourav Ghosh · Accepted Answer · 2016-01-15 14:34:17Z

1

In your case, ap[i][j] is allowed because, it is meaning ful.

Let's check the data types.

For int *ap[m];, ap is an array of int *s. In case of the function parameter int*ap[], ap is a pointer to a pointer-to-int.
Then, ap[k] (referring to the previous point) is an int *. This may very well be allocated memory which can provide valid access to more that one ints.
Provided enough memory allocated, ap[k][s] will refer to an int.

answered Jan 15, 2016 at 14:34

Sourav Ghosh

135k17 gold badges192 silver badges271 bronze badges

Comments

Iharob Al Asimi · Accepted Answer · 2016-01-15 14:32:10Z

1

Because you can dereference the pointer with index notation. First you access the element (the pointer) with an index, now the pointer can be dereferenced too with an index.

The equivalency between that and the indirection operator is as follows

pointer[i] == *(pointer + i);

answered Jan 15, 2016 at 14:32

Iharob Al Asimi

53.1k6 gold badges66 silver badges100 bronze badges

1 Comment

A.Lymphater Over a year ago

So the second index access the element the pointer is pointing to?

haccks · Accepted Answer · 2016-01-15 14:33:20Z

0

Both of the parameters ap and bp of function dosomethingwithmatrix are pointer to pointer to int. They are not array of pointers. The function declarator

int* dosomethingwithmatrix(const int *ap[], int *bp[])

is equivalent to

int* dosomethingwithmatrix(const int **ap, int **bp)

answered Jan 15, 2016 at 14:33

haccks

106k27 gold badges181 silver badges273 bronze badges

Collectives™ on Stack Overflow

access to an array of pointers in c

4 Answers 4

Comments

Comments

1 Comment

Comments

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

1 Comment

Comments

Related