In the just previous post: > It actually would be guaranteed to 0 it if it were instead unsigned chars. and bringing up padding. But this is padding within the

In the just previous post: > It actually would be guaranteed to 0 it if it were instead unsigned chars. and bringing up padding. But this is padding within the char, not padding within the struct. It is because of the latter, that sum(sizeof(members)) is not necessarily == to sizeof(struct). The sizeof(char) and sizeof(unsigned char) are both defined as 1, and of course the bit size of an unsigned char including padding is CHAR_BIT. And it turns out because of a char* must be able to access at least every accessible char of every other object, and have the weakest alignment that alignof(char) pretty much has to be 1 except in a contrived example http://port70.net/~nsz/c/c11/n1570.html#6.2.8 http://port70.net/~nsz/c/c11/n1570.html#6.2.8. Further, although not required, struct member padding will be a direct result of alignof, hence you will pretty much never find a wild example where sizeof(char)*2 == 2 != sizeof(struct a)... so maybe it is fair to call it a guarantee?? But, still, I think the way you're saying it just confuses the alignment/padding issue further.