Question

C Programming

Given the declarations:
struct STRUCTURE
{
int aa[7];
double dd;
char ff[30];
}
union Group
{
char bb[7];
int a[7];
}
struct STRUCTURE yy;
union Group zz;

(you may assume that sizeof(char)=1; sizeof(int)=4; sizeof(double)=8;

Number of bytes reserved for yy is

Number of bytes reserved for zz is

Answer 1

The memory is read in a block, and a block is of 8 bytes. When the variables are assigned and the memory is allocated, the memory allocation is done in such a way that 8 bytes are always read at a time. Hence the compiler leaves some useless space so that the provided variables can be fully fitted in a block size(i.e 8 bytes).
Hence the memory allocation will be
For int aa[7] = 7*4= 28 = 32(leaving some extra bytes to maintain block read write)
dd = 8
And ff = 30 = 32 (leaving some extra bytes to maintain block read write)

Hence total = 32+8+32 = 72 This is called structure padding. Therefore size of yy=72

For union, the total memory size of the union = size of biggest data variable, in this case it is int a[7] = 7*4 = 28 bytes. Therefore size of zz=28 bytes.