In Swift, why is "let _ = this" faster then "this != nil"? - ios

Question

So my question is why is let _ = this faster then this != nil?
Example:
This is:
let this : Bool? = true //
let start = DispatchTime.now()
for _ in 0...100000000 {
guard this != nil else { continue }
}
let end = DispatchTime.now()
let nanoTime = end.uptimeNanoseconds - start.uptimeNanoseconds
let timeInterval = Double(nanoTime)
print("Time \(timeInterval)")
// Time 5426559135.0
// Time 5428084767.0
// Time 5327325459.0
Slower than:
let this : Bool? = true //
let start = DispatchTime.now()
for _ in 0...100000000 {
guard let _ = this else { continue }
}
let end = DispatchTime.now()
let nanoTime = end.uptimeNanoseconds - start.uptimeNanoseconds
let timeInterval = Double(nanoTime)
print("Time \(timeInterval)")
// Time 257045414.0
// Time 261933863.0
// Time 263465919.0

Answer 1

Following Jonathan's response I checked the actual disassembled instructions.
Here is the result:
For the code:
let this : Bool? = nil
this != nil
we get:
0x100001290 <+0>: pushq %rbp
0x100001291 <+1>: movq %rsp, %rbp
0x100001294 <+4>: subq $0x30, %rsp
0x100001298 <+8>: leaq 0x2c7259(%rip), %rdx ; type metadata for Swift.Bool
0x10000129f <+15>: leaq 0x2b66ca(%rip), %rcx ; protocol witness table for Swift.Bool : Swift.Equatable in Swift
0x1000012a6 <+22>: leaq -0x18(%rbp), %rax
0x1000012aa <+26>: leaq -0x8(%rbp), %r8
0x1000012ae <+30>: movb $0x2, 0x2f940b(%rip)
0x1000012b5 <+37>: movb 0x2f9404(%rip), %r9b ; test2.this : Swift.Optional<Swift.Bool>
0x1000012bc <+44>: movb %r9b, -0x8(%rbp)
0x1000012c0 <+48>: movb $0x2, -0x10(%rbp)
0x1000012c4 <+52>: movb -0x10(%rbp), %r9b
0x1000012c8 <+56>: movb %r9b, -0x18(%rbp)
0x1000012cc <+60>: movl %edi, -0x1c(%rbp)
0x1000012cf <+63>: movq %r8, %rdi
0x1000012d2 <+66>: movq %rsi, -0x28(%rbp)
0x1000012d6 <+70>: movq %rax, %rsi
0x1000012d9 <+73>: callq 0x10004df10 ; Swift.!= infix <A where A: Swift.Equatable> (Swift.Optional<A>, Swift.Optional<A>) -> Swift.Bool
0x1000012de <+78>: xorl %r10d, %r10d
0x1000012e1 <+81>: movb %al, -0x29(%rbp)
0x1000012e4 <+84>: movl %r10d, %eax
0x1000012e7 <+87>: addq $0x30, %rsp
0x1000012eb <+91>: popq %rbp
0x1000012ec <+92>: retq
and for:
let this : Bool? = nil
let _ = this
there is:
0x1000012d0 <+0>: pushq %rbp
0x1000012d1 <+1>: movq %rsp, %rbp
0x1000012d4 <+4>: xorl %eax, %eax
0x1000012d6 <+6>: movb $0x2, 0x2f93e3(%rip)
0x1000012dd <+13>: movl %edi, -0x4(%rbp)
0x1000012e0 <+16>: movq %rsi, -0x10(%rbp)
0x1000012e4 <+20>: popq %rbp
0x1000012e5 <+21>: retq
Also, thank you Code Different for pointing to the Optimisation level.
Changing the value from [-Onone] to [-O -whole-module-optimisation] will cause a change for the generated asm in the following way:
The
let this : Bool? = nil
let _ = this
has
0x100001490 <+0>: pushq %rbp
0x100001491 <+1>: movq %rsp, %rbp
0x100001494 <+4>: movb $0x2, 0x3d9595(%rip) ; gCRAnnotations + 63
0x10000149b <+11>: xorl %eax, %eax
0x10000149d <+13>: popq %rbp
0x10000149e <+14>: retq
and the
let this : Bool? = nil
this != nil
to
0x100001490 <+0>: pushq %rbp
0x100001491 <+1>: movq %rsp, %rbp
0x100001494 <+4>: movb $0x2, 0x3d9595(%rip) ; gCRAnnotations + 63
0x10000149b <+11>: xorl %eax, %eax
0x10000149d <+13>: popq %rbp
0x10000149e <+14>: retq
So the resulted instructions are actually the same and the time to execute them should be pretty close.

Answer 2

I would check out this post. They both result in the same underlying assembly instructions. My guess is that they both take such a small amount of time to compile that the time difference you're noticing could be due to other miscellaneous outliers affecting performance.