When is memset to 0 nonportable?

From this comment in GCC bug #53119:

In C, {0} is the universal zero initializer equivalent to C++’s {} (the latter being invalid in C). It is necessary to use whenever you want a zero-initialized object of a complete but conceptually-opaque or implementation-defined type. The classic example in the C standard library is mbstate_t:

mbstate_t state = { 0 }; /* correctly zero-initialized */

versus the common but nonportable:

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mbstate_t state;
memset(&state, 0, sizeof state);

It strikes me as odd that the latter version could be unportable (even for implementation-defined types, the compiler has to know the size). What is the issue here and when is a memset(x, 0, sizeof x) unportable?

>Solution :

memset(p, 0, n) sets to all-bits 0.
An initializer of { 0 } sets to the value 0.
On just about any machine you’ve ever heard of, the two concepts are equivalent.

However, there have been machines where the floating-point value 0.0 was not represented by a bit pattern of all-bits-0. And there have been machines where a null pointer was not represented by a bit pattern of all-bits-0, either. On those machines, an initializer of { 0 } would always get you the zero initialization you wanted, while memset might not.

See also question 7.31 and question 5.17 in the C FAQ list.

Postscript: It’s not clear to me why mbstate_t would be a "classic example" of this issue, though.

P.P.S. One other difference, as pointed out by @ryker in a now-deleted answer: memset will set any "holes" in a padded structure to 0, while setting that structure to { 0 } might not.