Why is the size of this union larger than expected?

Question

When I type in the following code, I get a result of 32 bytes. It supposed to be 28 bytes, though. Why do I get that result?

//size of union variable

union U {
    int x[7];
    double y[3];
}z;

int main(){
    printf("%d",sizeof(z));
    return 0;
}

Answer 1

Unions, like structs, are padded to agree with the required alignment of their members.

double is aligned on a multiple of eight on your platform, and 32 is the smallest multiple of eight that is at least 28.

Answer 2

As explained in another answer, padding is used to facilitate struct (and union) member alignment and is the reason for the unexpected size that you see.

To see the size you expected (i.e. without padding), you can force the code to to ignore the alignment issue by using a pragma statement. #pragma pack specifically is used to indicate to the compiler that the struct/union be packed in a specific way, by changing the addressing alignment from the default value to something else.

The following results in a size of 28:

#pragma pack(1) //Sets compiler to align on 1 byte address boundaries
                //(Effectively removing all padding)

union U {
    int x[7];
    double y[3];
}z;

#pragma pack() //Set packing back to what it was before

int main(){
    printf("%d",sizeof(z));
    return 0;
}