Could somebody please explain the output of this code?
#include <iostream>
using namespace std;
struct Yo{
char sex;
int a;
};
int main() {
Yo c;
cout<<sizeof(c.sex);
cout<<endl<<sizeof(c.a);
cout<<endl<<sizeof(c);
return 0;
}
Output: 1 4 8
How is the size of structure 8?
This is memory alignment.
struct Yo{
char sex; // Takes up 1 byte + 3 for padding
int a; // Takes up 4 bytes
};
The three bytes between sex and a won't be used because the compiler aligns them for better performance. Thus sex ends up using 1 byte of space and the three bytes after the member variable are used as padding to ensure that int a has an address multiple of 4 (it is 4 byte aligned).
sex uses 4 bytes of space is wrong. The padding is not part of any member at all.
Because of structure padding (aka memory alignment). The alignment has to be a power of two (C11 makes this explicit in 6.2.8p4 as stated by @KeithThompson) and because the total size of your structure is 5, the nearest multiple of 4 is 8 so it gives 8, also because the alignment has to be a power of two.
You can use #pragma pack to have the exact size.
#pragma pack(push, 1) /* set alignment to 1 byte boundary */
struct A {
char s;
int a;
};
#pragma pack(pop) // restore to default
Warning: #pragma pack is not in standard C, nor is the assumption that the structure requires 4-byte alignment. As @KeithThompson stated.
"The size of a type must be a multiple of its alignment, but the size needn't be a power of 2. For example, assuming 4-byte int, a structure like struct { int a, b; char d; } will likely have an alignment of 4 and a size of 12. The size is the nearest multiple of the alignment, not the nearest power of 2." - @KeithThompson
Packing is useful to decrease memory, use it when you have a structure full of integers, fixed char lengths, etc. I do not know if it's good to use with pointers but I do not see it useful when using a pointer (e.g. a void * in the struct).
Read more here
#pragma pack()as said here at the end