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Is partial allocation of an object Undefined Behavior?
Is it valid to partly allocate an object, as long as you only use the allocated part of it?
#include <stdio.h>
#include <stdlib.h>
struct s {
int i[100];
};
int main(void)
{
struct s *s;
s = malloc(50 * sizeof(int));
s->i[30] = 7;
printf("%d\n", s->i[30]);
free(s);
}
alx@debian$ gcc --version | head -n1
gcc (Debian 12.2.0-13) 12.2.0
alx@debian$ clang --version | head -n1
Debian clang version 14.0.6
alx@debian$ clang -Weverything malloc.c -O3
alx@debian$ clang -Weverything malloc.c
alx@debian$ gcc -Wall -Wextra malloc.c -fanalyzer
alx@debian$ gcc -Wall -Wextra malloc.c -fanalyzer -O3
alx@debian$ ./a.out
7
The compiler don't seem to complain. It seems to work.
I didn't find anything in the standard that makes this code Undefined Behavior. However, I still suspect of it: the last half of the array within the structure is not allocated. Is it well-defined?
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| alx | (no comment) | Jan 26, 2023 at 13:30 |
Since I don't think the C standard says anything explicitly about cases like this, it is probably undefined behavior, under the "not mentioned in the standard" variety. If something isn't mentioned, it is per definition undefined and not well-defined.
The closest thing might be the somewhat unclear rules about "effective type" and "strict aliasing" in 6.5: "For all other accesses to an object having no declared type, the effective type of the object is simply the type of the lvalue used for the access." Everything returned from malloc has no declared type.
This doesn't mention anything about arrays, except that we can make an access through an "aggregate" (array or struct) that contains the effective type as one of its members.
However, nothing in the standard mentions what happens when you allocate too little space. As far as malloc is concerned, you have allocated an object with size 50 * sizeof(int) bytes, which has no declared type.
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