这取决于编译器。

如果它优化了代码，或者对于一个特定的指令集，如果它用相同数量的指令计算1和2为真，那么执行速度将是相同的。

在实际情况下，它总是同样快，但可以想象在特定的编译器和特定的系统中，这将得到不同的计算。

我的意思是:这不是一个与语言(C)相关的问题。

You should have asked him how did he reached to that conclusion. Under any decent compiler out there, the two compile to the same asm instructions. So, he should have told you the compiler as well to start off. And even so, you would have to know the compiler and platform very well to even make a theoretical educated guess. And in the end, it doesn't really matter in practice, since there are other external factors like memory fragmentation or system load that will influence the loop more than this detail.

根据人们花费在测试、证明和回答这个非常直接的问题上的时间和精力来判断，我认为这两者都因为提出这个问题而变得非常缓慢。

所以花更多的时间在这上面…

虽然(2)很荒谬，因为，

While(1)和While (true)在历史上被用于创建一个无限循环，该循环预计在循环中的某个阶段根据一定会发生的条件调用break。

1只是为了总是求值为真，因此，说while(2)和说while(1 + 1 == 2)一样愚蠢，后者也会求值为真。

如果你想完全傻一点，就用:-

while (1 + 5 - 2 - (1 * 3) == 0.5 - 4 + ((9 * 2) / 4.0)) {
    if (succeed())
        break;
}

我认为面试官犯了一个错别字，这并不影响代码的运行，但如果他故意使用2只是为了表现得奇怪，那么在他把奇怪的语句贯穿你的代码，使其难以阅读和使用之前解雇他。

这两个循环都是无限的，但我们可以看到哪个循环每次迭代需要更多的指令/资源。

使用gcc，我编译了以下两个程序，以不同的优化级别进行汇编:

int main(void) {
    while(1) {}
    return 0;
}

int main(void) {
    while(2) {}
    return 0;
}

即使没有优化(-O0)，两个程序生成的程序集也是相同的。因此，两个循环之间没有速度差异。

作为参考，下面是生成的程序集(使用gcc main.c -S -masm=intel并带有优化标志):

o0:

    .file   "main.c"
    .intel_syntax noprefix
    .def    __main; .scl    2;  .type   32; .endef
    .text
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    push    rbp
    .seh_pushreg    rbp
    mov rbp, rsp
    .seh_setframe   rbp, 0
    sub rsp, 32
    .seh_stackalloc 32
    .seh_endprologue
    call    __main
.L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

o1群:

    .file   "main.c"
    .intel_syntax noprefix
    .def    __main; .scl    2;  .type   32; .endef
    .text
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    sub rsp, 40
    .seh_stackalloc 40
    .seh_endprologue
    call    __main
.L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

-O2和-O3(相同输出):

    .file   "main.c"
    .intel_syntax noprefix
    .def    __main; .scl    2;  .type   32; .endef
    .section    .text.startup,"x"
    .p2align 4,,15
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    sub rsp, 40
    .seh_stackalloc 40
    .seh_endprologue
    call    __main
.L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

事实上，为循环生成的程序集对于每个级别的优化都是相同的:

 .L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

重要的部分是:

.L2:
    jmp .L2

我不太懂汇编，但这显然是一个无条件循环。jmp指令无条件地将程序重置回. l2标签，甚至不将值与true进行比较，当然，它会立即再次这样做，直到程序以某种方式结束。这直接对应于C/ c++代码:

L2:
    goto L2;

编辑:

有趣的是，即使没有优化，下面的循环都在汇编中产生完全相同的输出(无条件jmp):

while(42) {}

while(1==1) {}

while(2==2) {}

while(4<7) {}

while(3==3 && 4==4) {}

while(8-9 < 0) {}

while(4.3 * 3e4 >= 2 << 6) {}

while(-0.1 + 02) {}

甚至令我惊讶的是:

#include<math.h>

while(sqrt(7)) {}

while(hypot(3,4)) {}

对于用户定义函数，事情变得更加有趣:

int x(void) {
    return 1;
}

while(x()) {}

#include<math.h>

double x(void) {
    return sqrt(7);
}

while(x()) {}

At -O0, these two examples actually call x and perform a comparison for each iteration.

First example (returning 1):

.L4:
    call    x
    testl   %eax, %eax
    jne .L4
    movl    $0, %eax
    addq    $32, %rsp
    popq    %rbp
    ret
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

Second example (returning sqrt(7)):

.L4:
    call    x
    xorpd   %xmm1, %xmm1
    ucomisd %xmm1, %xmm0
    jp  .L4
    xorpd   %xmm1, %xmm1
    ucomisd %xmm1, %xmm0
    jne .L4
    movl    $0, %eax
    addq    $32, %rsp
    popq    %rbp
    ret
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

However, at -O1 and above, they both produce the same assembly as the previous examples (an unconditional jmp back to the preceding label).

TL;DR

Under GCC, the different loops are compiled to identical assembly. The compiler evaluates the constant values and doesn't bother performing any actual comparison.

The moral of the story is:

• There exists a layer of translation between C source code and CPU instructions, and this layer has important implications for performance.
• Therefore, performance cannot be evaluated by only looking at source code.
• The compiler should be smart enough to optimize such trivial cases. Programmers should not waste their time thinking about them in the vast majority of cases.

我认为线索可以从“高级经理的询问”中找到。这个人显然在成为经理后就停止了编程，然后他/她花了几年时间才成为高级经理。从未失去对编程的兴趣，但从那以后再也没有写过一行字。所以他提到的并不是“任何一个像样的编译器”，而是“这个人在二三十年前合作过的编译器”。

At that time, programmers spent a considerable percentage of their time trying out various methods for making their code faster and more efficient as CPU time of 'the central minicomputer' was so valueable. As did people writing compilers. I'm guessing that the one-and-only compiler his company made available at that time optimized on the basis of 'frequently encountered statements that can be optimized' and took a bit of a shortcut when encountering a while(1) and evaluated everything else, including a while(2). Having had such an experience could explain his position and his confidence in it.

让你被录用的最好方法可能是，在你顺利地引导他进入下一个面试主题之前，让高级经理忘乎所以地给你讲2-3分钟“编程的美好时光”。(好的时机在这里很重要——太快你会打断故事——太慢你会被贴上焦点不集中的标签)。一定要在面试结束时告诉他，你对这个话题非常感兴趣。

如果你担心优化，你应该使用

for (;;)

因为它没有检验。(愤世嫉俗者模式)