5 ms·
In C it also helps to think of it in the equivalent pointer form: p == &(p[0]) p[0] == *(p+0) p[1] == *(p+1) where p is a pointer to some arbitrary d
by jlubawy 10y ago
In C it also helps to think of it in the equivalent pointer form:
p == &(p[0])
p[0] == *(p+0)
p[1] == *(p+1)
where p is a pointer to some arbitrary data type. Then it becomes an offset in memory.
- deleted 10y ago[deleted]