10.00.a. What would the command: printf(“%c”, mystring[6]); yield?

We know that mystring (== &mystring[0]) == 4060 (the base address of array mystring).

Therefore, &mystring[6] is: 4060 + 6 * 1 = 4066.

When we go to that location (4066), we find the bit pattern:

01110101

Evaluating:

2⁶ + 2⁵ + 2⁴ + 2² + 2⁰ = 64 + 32 + 16 + 4 + 1 = 117

Since ‘a’ is ASCII 97, ASCII 117 is the character ‘u’ (lower case u)