Swift 是否实现了尾调用优化?在相互递归的情况下? [英] Does Swift implement tail call optimization? and in mutual recursion case?

问题描述

特别是如果我有以下代码:

In particular if I have the following code:

func sum(n: Int, acc: Int) -> Int {
  if n == 0 { return acc }
  else { return sum(n - 1, acc + n) }
}

Swift 编译器会将其优化为循环吗?在下面一个更有趣的案例中是这样吗?

Will Swift compiler optimize it to a loop? And does it so in a more interesting case below?

func isOdd(n: Int) -> Bool {
  if n == 0 { return false; }
  else { return isEven(n - 1) }
}

func isEven(n: Int) -> Bool {
  if n == 0 { return true }
  else { return isOdd(n - 1) }
}

推荐答案

最好的检查方法是检查编译器生成的汇编语言代码.我把上面的代码编译成:

The best way to check is to examine the assembly language code generated by the compiler. I took the code above and compiled it with:

swift -O3 -S tco.swift >tco.asm

输出的相关部分

.globl    __TF3tco3sumFTSiSi_Si
    .align    4, 0x90
__TF3tco3sumFTSiSi_Si:
    pushq    %rbp
    movq    %rsp, %rbp
    testq    %rdi, %rdi
    je    LBB0_4
    .align    4, 0x90
LBB0_1:
    movq    %rdi, %rax
    decq    %rax
    jo    LBB0_5
    addq    %rdi, %rsi
    jo    LBB0_5
    testq    %rax, %rax
    movq    %rax, %rdi
    jne    LBB0_1
LBB0_4:
    movq    %rsi, %rax
    popq    %rbp
    retq
LBB0_5:
    ud2

    .globl    __TF3tco5isOddFSiSb
    .align    4, 0x90
__TF3tco5isOddFSiSb:
    pushq    %rbp
    movq    %rsp, %rbp
    testq    %rdi, %rdi
    je    LBB1_1
    decq    %rdi
    jo    LBB1_9
    movb    $1, %al
LBB1_5:
    testq    %rdi, %rdi
    je    LBB1_2
    decq    %rdi
    jo    LBB1_9
    testq    %rdi, %rdi
    je    LBB1_1
    decq    %rdi
    jno    LBB1_5
LBB1_9:
    ud2
LBB1_1:
    xorl    %eax, %eax
LBB1_2:
    popq    %rbp
    retq

    .globl    __TF3tco6isEvenFSiSb
    .align    4, 0x90
__TF3tco6isEvenFSiSb:
    pushq    %rbp
    movq    %rsp, %rbp
    movb    $1, %al
LBB2_1:
    testq    %rdi, %rdi
    je    LBB2_5
    decq    %rdi
    jo    LBB2_7
    testq    %rdi, %rdi
    je    LBB2_4
    decq    %rdi
    jno    LBB2_1
LBB2_7:
    ud2
LBB2_4:
    xorl    %eax, %eax
LBB2_5:
    popq    %rbp
    retq

生成的代码中没有调用指令，只有条件跳转(je/jne/jo/jno).这清楚地表明 Swift 确实在两种情况下都做了尾调用优化.

There are no call instructions in the generated code, only conditional jumps (je / jne / jo / jno). This clearly suggests that Swift does do tail call optimizations in both cases.

此外，isOdd/isEven 函数很有趣，因为编译器似乎不仅执行 TCO，而且在每种情况下内联其他函数.

In addition, the isOdd/isEven functions are interesting in that the compiler not only seems to perform TCO but also inlines the other function in each case.

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