Pointer-to-array, malloc, and out-of-bounds access

Question

Given a pointer-to-array in C, can we malloc to it enough memory for extra elements (beyond the specified array size) and then safely access those elements using either the [] operator or pointer arithmetic?

Consider this example:

int (*foo)[ 10 ]; //Declare pointer to array of ten ints
foo = malloc( sizeof( int ) * 20 ); //Allocate memory for TWENTY ints
(*foo)[ 15 ] = 100; //Access element in "extra" space via [] operator 
*( *foo + 16 ) = 200; //Access element in "extra" space via pointer arithmetic
printf( "%d\n", *( *foo + 15 ) ); //Prints 100
printf( "%d\n", (*foo)[ 16 ] ); //Prints 200

This code compiles fine and produces correct results in gcc. However, I'm not sure if it invokes undefined behavior.

Thanks!

foo[x] is the same as *(foo + x).

Marco
– Marco

2021-10-31 01:57:04 +00:00
Commented Oct 31, 2021 at 1:57 — Marco
– Marco, Commented Oct 31, 2021 at 1:57

dbush · Accepted Answer · 2021-10-31 00:39:33Z

2

What you're doing trigger undefined behavior because you're reading/writing past the bounds of an array of int of size 10.

The proper way to access this memory is to use 2D array access. So instead of this:

(*foo)[15] = 100;

Which is equivalent to this:

foo[0][15] = 100;

Do this:

foo[1][5] = 100;

answered Oct 31, 2021 at 0:39

dbush

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3 Comments

Jackson Allan Over a year ago

Thanks for the response. Are you sure the out-of-bounds access rule applies to the pointer arithmetic version?: ( *foo + n ). *foo + 0 decays to int pointer pointing the first element in the array. However, it's also an int pointer to the beginning of a malloced bloc of memory, which the the standards seem to treat as an "array object" and which I should - in theory - be able to move around inside freely using pointer arithmetic.

Jackson Allan Over a year ago

As for the proper/usual way, it's not quite relevant because the idea I'm exploring here is that pointers-to-array could be used to tack extra, compile-time information (specifically an integer value) onto what would otherwise be a regular pointer.

dbush Over a year ago

@JacksonAllan The syntax E1[E2] is exactly equivalent to *(E1 + E2) so yes it applies. As for arithmetic into the allocated memory, if you assigned the malloced memory to a char *, do arithmetic on that, then convert the resulting pointer to the desired type, that would probably be OK since a char * gets special treatment.

0___________ · Accepted Answer · 2021-10-31 01:20:32Z

0

I would allocate it a bit different way:

foo = malloc( sizeof( *foo ) * nrows );

it will allocate 2D array having 10 clomns and nrows rowas.

The best way to access the array is using indexes

foo[row][column]

answered Oct 31, 2021 at 1:20

0___________

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