Does struct assignment guarantee padding to also be equal

Question

Say I have a struct with 2 fields and the implementation of C that I have also has some padding between these fields.

If I create two variables of the struct and assign one to another, will the padding be guaranteed to be equal?

I know that for most compilers it would be so (because they just call memcpy), but I want to know what is specified about the padding in the standard?

Intention for this question is, can I use memcmp to check equality of structs.

Say I have a compiler which emits code that just assigns all the members of the struct instead of doing memcpy, will it be a valid implementation of the assignment of struct operation?

Answer 1

6.2.6.1 General
...
6    When a value is stored in an object of structure or union type, including in a member object, the bytes of the object representation that correspond to any padding bytes take unspecified values.⁵¹⁾ The value of a structure or union object is never a trap representation, even though the value of a member of the structure or union object may be a trap representation.

7    When a value is stored in a member of an object of union type, the bytes of the object representation that do not correspond to that member but do correspond to other members take unspecified values.

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^{51) Thus, for example, structure assignment need not copy any padding bits.}

^{C 2011 Online Draft}

Footnote 51 directly addresses your question - contents of padding bits may not be copied over in assignment, so memcmp may not work for comparing two structs for equality.

Answer 2

The standard says in a note 51 of 6.2.6.1 General:

Thus, for example, structure assignment need not copy any padding bits.

Answer 3

I don't think that you can use memcmp, because the memory between the members (i.e. the content in the "padding"-area) is not guaranteed to have a certain value, and assigning such a struct-object to another is not required to copy the padding content as well.