Memory

Padding

struct Garbage { 
        double a;
        int b;
        short c;
        void *c;
        char e;
        float *f;
        char g;
}

This will result in the following padding

8 double
4 int  
2 short   
2 paddding
8 pointer
1 char   
7 padding
8 pointer
1 char   
7 padding

Thus there is a 2 + 7 + 7 = 16 bit of padding that is wasted.

• Let \(n=2^x\) for some non-negative integer \(x\), and let \(a\) denote a memory address
• If a % n == 0, then we are done since \(a\) is alligned
• Otherwise, construct a bitmask, \(b\), of \(|a|\) bits with the value \(n - 1\).
• TODO