1

Suppose i have two dimentional array like this:

A[2][3] = {{1,2,4},{2,5,15}}

and if i were to pass it as a function argument 

#include <stdio.h>

int fun(int A[][3])  // or int fun( int (*A)[3])
{
.
.
}

main()
{
int A[2][3] = {{1,2,4},{2,5,15}}; 
fun(A); 
}

Q) Why is this first block can be empty in function argument like this [] and it can stay empty why can't others stay empty as well like A[][] ? Or why this is wrong **A as a function argument?

Now if i have a three dimentional array like this:

A[2][3][4] = {{{2,21,12,34,23},{322,34,34,23},{323,32,112,324}}, {{23,231,21,233},{23,5,65,2},{3,23,234,34}}}

and if i were to pass it as a function argument 

#include <stdio.h>

int fun(int A[][3][4])  // or int fun( int (*A)[3][4])
{
.
.
}

main()
{
int A[2][3][4] = {{{2,21,12,34,23},{322,34,34,23},{323,32,112,324}}, {{23,231,21,233},{23,5,65,2},{3,23,234,34}}}

fun(A); 
}

Q) Why is this first block can be empty in function argument like this [] and it can stay empty why can't others stay empty as well like A[][][] ? Or why this is wrong ***A as a function argument?

HELP PLZ
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  • Why can code pass `char s[6] = "Hello";` to `puts(const char *s);` without specifying the array size of `s`? Same reason. – chux - Reinstate Monica Aug 11 '14 at 23:29
  • Addressing offsetting. – Fiddling Bits Aug 11 '14 at 23:31
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    Top size (leftmost) than is required to calculate the address of the element as an array. The size of the leftmost is not required since it is not checked in the C element is calculated conversely whether they fall within the range of the array. **A -> *A and ***A -> **A What you dereference of these is a pointer but the pointer is not present there in the actual sequence. – BLUEPIXY Aug 11 '14 at 23:31

2 Answers2

3

The Standard says:

An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to an rvalue of type “pointer to T.” The result is a pointer to the first element of the array.

Simply put, it allows argument int A[x][y][z] to be converted to int (A*)[y][z]. We usually say that such an array decays to a pointer.

The Standard also says:

Parameter declarations that differ only in a pointer * versus an array [] are equivalent. That is, the array declaration is adjusted to become a pointer declaration (8.3.5). Only the second and subsequent array dimensions are significant in parameter types (8.3.4).

Which means that int (A*)[y][z] is no difference from int A[][y][z] when used as function parameter.

Thus, it's ok to just pass int A[x][y][z] (but not int A[][][z] or int A[][][], etc) to a function which requires a parameter of type int A[][y][z].

Why size of low dimensions are required? Because compilers need to know the type of element A* points to (here is int [y][z]) so that it can correctly do pointer arithmetic. E.g., A's value = A's value + sizeof(int) * y * z when you ++A.

Eric Z
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1

You can pass a pointer to an arbitrarily dimensioned array as an argument, but if you expect the compiler to do pointer arithmetic using it, the compiler will need to know the size of the various dimensions. See Passing multidimensional arrays as function arguments in C for more explanation.

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Fred
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