嵌套if语句和“&&”操作者 [英] Nested if statements and "&&" operator

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本文介绍了嵌套if语句和“&&”操作者的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

if(a() && b() && c() && d())
   doSomething();


if(a())
   if(b()) 
      if(c())
         if(d())
            doSomething();

这两者之间是否存在任何性能差异?

Is there "any" performance difference between these two?

例如,在a()变为0的情况下,它会在第一个if语句中继续运行b(),c()和d()吗?或者它与第二个嵌套的if语句一样吗?

For example, in a situation that a() turns 0, will it keep running b(), c() and d() in the first if statement? Or will it work same as the second nested if statement?

推荐答案

它们完全相同。

要自己测试一下,运行 gcc -S test.c (假设这是您放置源代码的地方)并观察内容 test.s

To test this yourself, run gcc -S test.c (presuming that this is where you've put your source) and observe the contents of test.s.

以下是嵌套的方法 - 如果方法在gcc 4.8.1中使用默认选项编译(带注释注释):

Here's how the nested-if approach compiles in gcc 4.8.1 with default options (annotated with comments):

main:
.LFB0:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    movl    $0, %eax
    call    A                        # try to call A
    testl   %eax, %eax               # look at its return value
    je  .L3                          # short-circuit if it returned 0
    movl    $0, %eax                 # ...repeat for B, et al.
    call    B
    testl   %eax, %eax
    je  .L3
    movl    $0, %eax
    call    C
    testl   %eax, %eax
    je  .L3
    movl    $0, %eax
    call    D
    testl   %eax, %eax
    je  .L3
    movl    $0, %eax
    call    doSomething
.L3:
    popq    %rbp
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc

以下是&& 方法的编制方式:

Here's how the && approach compiles:

main:
.LFB0:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    movl    $0, %eax
    call    A                           # try to call A
    testl   %eax, %eax                  # look at its return value
    je  .L3                             # short-circuit if it returned 0
    movl    $0, %eax                    # ...repeat for B, et al.
    call    B
    testl   %eax, %eax
    je  .L3
    movl    $0, %eax
    call    C
    testl   %eax, %eax
    je  .L3
    movl    $0, %eax
    call    D
    testl   %eax, %eax
    je  .L3
    movl    $0, %eax
    call    doSomething
.L3:
    popq    %rbp
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc

这篇关于嵌套if语句和“&&”操作者的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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