Alignment when using a long double in a structure

Question

I know in general, a struct instance will have the alignment of its widest scalar member. I declared a structure having a member of long double data type.

struct try
{
    char a;
    long double b;
};
struct try obj;

When i tried to check sizeof(obj) it is coming out as 16. My compiler assumes long double as 12 bytes. So i am not able to understand how exactly padding is being done here and how alignment is happening in structure. I assumed that alignment will be done on basis of long double as it is the widest scalar member. So there should be a 11 byte padding for char and the size of structure variable should come out as 24 but output is 16. So exactly what is happening here ?. I am working on a 64 bit processor.

Apparently alignment is on `int` 32 size: one int for `a` is 4 bytes plus 12 for the long float makes 16. — Paul Ogilvie, Apr 25 '20 at 17:23
A structure doesn't always have to be aligned to its size. For example, a `char[100]` has size 100 but alignment 1. — Joseph Sible-Reinstate Monica, Apr 25 '20 at 17:23
@JosephSible-ReinstateMonica, the variable will start on an int boundary. — Paul Ogilvie, Apr 25 '20 at 17:24
@PaulOgilvie GCC and clang both tell me `_Alignof(char[100])` is 1. — Joseph Sible-Reinstate Monica, Apr 25 '20 at 17:25
@JosephSible-ReinstateMonica, so `char a[101], char b` would be contiguous? — Paul Ogilvie, Apr 25 '20 at 17:26
@PaulOgilvie Yes: https://godbolt.org/z/f7K3FA (Or in a struct: https://godbolt.org/z/NQw85b) — Joseph Sible-Reinstate Monica, Apr 25 '20 at 17:28
@S.S.Anne: https://gcc.gnu.org/onlinedocs/gcc/x86-Options.html#index-m96bit-long-double With -m32, it's 12 bytes and with -m64 it's 16 (by default) — rici, Apr 25 '20 at 17:35
IIRC, MSVC makes long double the same as double, so there it's 8 bytes. — rici, Apr 25 '20 at 17:36
@Noshiii Run this program on your compiler and see what it prints: https://godbolt.org/z/NQqhPv — Joseph Sible-Reinstate Monica, Apr 25 '20 at 17:42
@PaulOgilvie: I don't think it's guaranteed, but there's no need to pad and GCC doesn't. https://tio.run/##ZcixCsIwEADQ3a84KkICVewc8UfUIbnk7EFNJLk4WPrtsbiJ2@Ph/o7Ythxxqj7AqYj3gQ7jefNznP5qYrde4yjwsBzVK7HX8zOvQarbves1dj0kohIkkSqSK8qMo81gL8NxuJmvnVl6cFqbpbUP — rici, Apr 25 '20 at 17:42
@JosephSible-ReinstateMonica: https://tio.run/##Zc69DgIhEATg3qfYnDGBhLPQTo2Jz2Brw/HnJrAYAYszvrqI1@mV800xo3qnVF0iKV@0gUPKWnp0tL4eFz@qjZ0Zxhl5HJpVpAxBIrFHRM2ft3sDy7ozjmYHq7EIOH1XgqE85Qt1AlJrowXmIznQsQzecAHTnX/m@1etb2W9dKn2Ybv5AA (Note the -m32 flag) — rici, Apr 25 '20 at 17:47

score 4 · Accepted Answer · answered Apr 25 '20 at 18:05

4

First of all, your compiler is creating 32-bit output even though you have a 64-bit processor. Anyway, you're assuming that things need to be aligned to a boundary identical to their size, which isn't true in general. In particular, in your case, long doubles take up 12 bytes, but only need to be aligned to a 4-byte boundary. As such, after your single-byte char, the compiler inserts 3 bytes of padding to get to a 4-byte boundary, and then inserts your 12-byte long double. 1+3+12 is 16, so struct try is 16 bytes long.

answered Apr 25 '20 at 18:05

Joseph Sible-Reinstate Monica

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I am following this article for padding and align. : http://www.catb.org/esr/structure-packing/#_structure_alignment_and_padding . Whenever i use a datatype which is less than or equal to 8 bytes in size this rule is working properly(all structure elements are getting aligned to the widest scalar member). but you said that long double is needed to be aligned to 4 byte boundary only. what is reason behind this ? I have'nt read any such rule in the article above. Can you please explain more deeply how actually boundary alignments take place especially when data type is more than 8 byte of size. – LocalHost Apr 25 '20 at 18:18
That doesn't contradict what I wrote. There are two scalar members in your struct: `char`, with an alignment of 1, and `long double`, with an alignment of 4. The widest of 1 and 4 is 4, and your structure indeed has alignment 4. – Joseph Sible-Reinstate Monica Apr 25 '20 at 18:20
but as i said `long double` is of 12 bytes in my compiler. That's what my whole question is about. Why you are using 4 bytes for alignment. There gotta be some logic behind that ? sorry if i am missing something ? – LocalHost Apr 25 '20 at 18:32
@Noshiii That's just what the i386 ABI says. In your compiler, see what `_Alignof(long double)` is. It'll be 4. – Joseph Sible-Reinstate Monica Apr 25 '20 at 18:33
Yes it's 4. I got it. thanks a lot. Just one last thing, Alignment of structure depends on `_alignof(type)` and not on `sizeof()` ? Am i right ?? – LocalHost Apr 25 '20 at 18:43
1

Yes, you're right. Note, though, that it's either `alignof` (from `stdalign.h`) or `_Alignof` (the built-in). `_alignof` doesn't exist. – Joseph Sible-Reinstate Monica Apr 25 '20 at 18:45
oh okay, my bad, might be a typo. Will remember in future. thanks a lot for help. – LocalHost Apr 25 '20 at 18:47

Alignment when using a long double in a structure

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