gcc: how is "implementation-defined" malloc(0) defined?

Question

C standard (e.g. C99) tells that malloc(0) returns "a null pointer or a unique pointer that can be successfully passed to free()". Which of the two, is implementation-defined. (This has been asked on StackOverflow many times, e.g. 1, 2, 3, 4.)

My question is: how does GCC define it?

As whatever the C library on the target system provides?

Marc Glisse
– Marc Glisse

2015-01-14 16:31:19 +00:00
Commented Jan 14, 2015 at 16:31 — Marc Glisse
– Marc Glisse, Commented Jan 14, 2015 at 16:31

Dietrich Epp · Accepted Answer · 2015-01-14 16:31:33Z

5

GCC doesn't define it at all. GCC uses the malloc() implementation provided by the standard library. On OS X this will be libSystem, on Linux this will usually be glibc or eglibc, etc. These libraries are open source, so you can browse the source code.

answered Jan 14, 2015 at 16:31

Dietrich Epp

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1 Comment

Konstantin Shemyak Over a year ago

@Dietrich: Thank you! bazaar.launchpad.net/~vcs-imports/glibc/master/view/head:/… says that "Even a request for zero bytes (i.e., malloc(0)) returns a pointer to something of the minimum allocatable size."

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