setenv ("NAME", NULL) corrupts environment

Created attachment 5948
Proposed patch, which treats NULL value as "" (which I think is what the current code intends).

setenv ("NAME", NULL) corrupts the environment. It doesn't seem specified what the function should do when value is NULL, but the code does check for it - it just does the wrong thing:

stdlib/setenv.c:
...
__add_to_environ (name, value, combined, replace)
...
  const size_t vallen = value != NULL ? strlen (value) + 1 : 0;
...
      memcpy (new_value, name, namelen);
      new_value[namelen] = '=';
      memcpy (&new_value[namelen + 1], value, vallen);
...

i.e. the new value is set to "NAME=" without the trailing nul character.

The parameter is supposed to be a string pointer. NULL is no string pointer, neither is in most case -1 etc. There is no way to catch invalid pointers and any such effort would only hurt correct code. Just fix your code.

Could you please consider changing this line then:
  const size_t vallen = value != NULL ? strlen (value) + 1 : 0;
to:
  const size_t vallen = strlen (value) + 1;

This is detecting that value is NULL, handling it, then causing a greater problem later on in the function. If value must be non NULL then a segmentation fault is preferable to a memory corruption which is much harded to diagnose.

(In reply to comment #2)
> Could you please consider changing this line then:

No, this would require unnecessary other changes.

(In reply to comment #3)
> (In reply to comment #2)
> > Could you please consider changing this line then:
>
> No, this would require unnecessary other changes.

Since "value" and "combined" cannot be non-NULL altogether, the change is not going to be so tremendous:

- const size_t vallen = value != NULL ? strlen (value) + 1 : 0;
+ const size_t vallen = combined == NULL ? strlen (value) + 1 : 0;