This design keeps arithmetic in registers. What touches memory?
- Only loads and stores
- Every arithmetic operation
- Only the compiler
What does an instruction set fix?
- How fast the chip runs
- Which operations exist, which registers are visible, and how memory is addressed
- How big the power supply is
Two chips with different internals can run the same binary if both keep the contract.
Circle one: True False
A chip schedules work differently inside but keeps the same visible operations and registers. Will old programs still run?
- Yes, because programs only see the contract
- No, because the inside changed
- Only after recompiling
Compared with load-store, what does operating directly on memory cost?
- Programs need no registers at all
- Memory becomes as fast as registers
- Each operation grows more complex, while loads and stores are saved
Which change keeps the promise to old programs?
- Renaming a visible register
- Changing how work is scheduled inside the chip
- Removing an operation compilers use
What does the instruction set promise a compiler?
- Every chip will run at the same speed
- These operations and registers will be there on every chip that keeps the contract
- No chip will ever gain new internals
A designer renames a visible register and says nothing breaks. What is wrong?
- The schedule inside runs slower
- The chip uses more power
- Programs address that register, so the contract changed