Which is faster: x<<1 or x<<10?

Potentially depends on the CPU.

However, all modern CPUs (x86, ARM) use a "barrel shifter" -- a hardware module specifically designed to perform arbitrary shifts in constant time.

So the bottom line is... no. No difference.

Some embedded processors only have a "shift-by-one" instruction. On such processors, the compiler would change x << 3 into ((x << 1) << 1) << 1.

I think the Motorola MC68HCxx was one of the more popular families with this limitation. Fortunately, such architectures are now quite rare, most now include a barrel shifter with a variable shift size.

The Intel 8051, which has many modern derivatives, also cannot shift an arbitrary number of bits.

There are many cases on this.

1. Many hi-speed MPUs have barrel shifter, multiplexer-like electronic circuit which do any shift in constant time.

2. If MPU have only 1 bit shift x << 10 would normally be slower, as it mostly done by 10 shifts or byte copying with 2 shifts.

3. But there is known common case where x << 10 would be even faster than x << 1. If x is 16 bit, only lower 6 bits of it is care (all other will be shifted out), so MPU need to load only lower byte, thus make only single access cycle to 8-bit memory, while x << 10 need two access cycles. If access cycle is slower than shift (and clear lower byte), x << 10 will be faster. This may apply to microcontrollers with fast onboard program ROM while accessing slow external data RAM.

4. As addition to case 3, compiler may care about number of significant bits in x << 10 and optimize further operations to lower-width ones, like replacing 16x16 multiplication with 16x8 one (as lower byte is always zero).

Note, some microcontrollers have no shift-left instruction at all, they use add x,x instead.