在 for 循环中迭代所有无符号整数 [英] Iterating over all unsigned integers in a for loop

问题描述

假设我想在 for 循环中遍历所有整数.为了便于讨论，假设我正在为每个整数调用一些未知函数 f(unsigned x):

Let's say I want to iterate over all integers in a for loop. For the sake of discussion, assume I am calling some unknown function f(unsigned x) for each integer:

for (unsigned i = 0; i < UINT_MAX; i++) {
     f(i);
}

当然，上面没有遍历所有整数，因为它错过了一个:UINT_MAX.将条件更改为 i <= UINT_MAX 只会导致无限循环，因为这是同义反复.

Of course, the above fails to iterate over all integers, because it misses one: UINT_MAX. Changing the condition to i <= UINT_MAX just results in an infinite loop, because that's a tautology.

您可以使用 do-while 循环来实现，但是您会失去 for 语法的所有细节.

You can do it with a do-while loop, but you lose all the niceties of the for syntax.

我可以吃蛋糕(for 循环)并吃掉它(迭代所有整数)吗?

Can I have my cake (for loops) and eat it too (iterate over all integers)?

推荐答案

你可以用 do-while 循环来做，但是你失去了所有的细节for 语法.

You can do it with a do-while loop, but you lose all the niceties of the for syntax.

通过使用匿名块作用域，它仍然可以使用 do-while 循环:

It is still doable with do-while loop by using an anonymous block scope:

{
    unsigned i = 0;
    do { f(i); } while (++i != 0);
}

虽然这个结构可能不是最惯用的，但它显然是清晰的汇编代码的候选者.例如，gcc -O 将其编译为:

While this construct may not be most idiomatic, it is an obvious candidate for clear assembly code. For example, gcc -O compiles it as:

.L2:
        mov     edi, ebx   ; ebx starts with zero
        call    f
        add     rbx, 1
        cmp     rbx, rbp   ; rbp is set with 4294967296
        jne     .L2

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