嵌套 if 语句和“&"操作员

Nested if statements and quot;amp;amp;quot; operator(嵌套 if 语句和“amp;操作员)

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?或者它会和第二个嵌套的 if 语句一样工作吗?

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，请运行 gcc -S test.c(假设这是您放置源代码的位置)并观察 test.s 的内容.

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.

以下是使用默认选项(用注释注释)在 gcc 4.8.1 中编译nested-if 方法的方式:

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 && 方法的编译方式:

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

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