为什么按位左移将uint8_t提升为更宽的类型 [英] Why does bitwise left shift promotes an uint8_t to a wider type

问题描述

我对uint8_t有点困惑，很好奇当我向左流出位并发现

I was a bit messing around with uint8_t and was curious what happens when I outflow bits to the left and found that

uint8_t i = 234;

uint8_t j = (i << 1);
auto    k = (i << 1);

std::cout << (int)j << std::endl;

std::cout << k << std::endl;

打印出

212
468

而不是预期的

212
212

似乎<<确实将uint8_t提升了一些更宽的整数类型.为什么这样做呢?

It seems like << does promote an uint8_t too some wider integer type. Why does it do this?

这里一个链接，你看到它在行动

Here a link where you see it in action

推荐答案

几乎每个算术运算都执行所谓的常规算术转换.

Pretty much every arithmetic operation performs what's called the usual arithmetic conversions.

这可以追溯到几十年前.

This goes back decades.

首先，进行整体促销.

• 没有算术运算占用uint8_t，因此两个操作数将始终被提升.

• No arithmetic operation takes uint8_t so both of your operands will always be promoted.

此后，找到一个通用类型，并在必要时进行转换.

After that, a common type is found and conversions take place if necessary.

• 您可以通过将右侧的类型强制转换为i的类型来防止这种情况，但是根据上述规则，在这种情况下您什么都不会得到.

• You can prevent this by casting the right-hand-side to the type of i but, per the above rule, that doesn't get you anything in this case.

(您可以在此处和结果是表达式的结果永远不会是uint8_t；只是在j的情况下，您已将其 back 强制转换为uint8_t，并随之产生了环绕.

The upshot is that the result of your expression is never going to be uint8_t; it's just that in the case of j you've cast it back to uint8_t, with the wraparound that consequently ensues.

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