解决复合赋值转换警告 [英] Resolving a conversion warning in a compound assignment

问题描述

在我的code我有很多的变量＆LT;＆LT; = 1; 的句子，其中变量的类型为uint16_t的。编译器随地吐痰警告说

In my code I have a lot of variable <<= 1; sentences where variable is of type uint16_t. The compiler is spitting a warning saying

转换从'诠释'uint16_t可能会改变其值[-Wconversion]

conversion to 'uint16_t' from 'int' may alter its value [-Wconversion]

我该如何解决呢？我可以用一个长形的符号，比如变量=（uint16_t）（可变＆LT;＆LT; 1） - 但我想保持短期符号。

How can I resolve it? I could use a long-form notation like variable = (uint16_t)(variable << 1) - but I'd like to keep the short notation.

推荐答案

根据我的标准的阅读，你不能从这个警告这个原因离开：

Based upon my reading of the standard, you can not get away from this warning for this reason:

uint16_t foo;
foo <<= 1;

等同于

uint16_t foo;
foo = foo << 1;

不过，这是在整型提升天下赶上了。

However, this is caught up in the world of "integer promotion".

在富＆LT;＆LT; 1 的值前pression有类型富，然而之前可以进行左移它首先必须通过整数推广; C99标准部分的 6.3.1.1.2 规定：如果int可以重新present原始类型的所有值，该值被转换为int

The value of the "foo << 1" expression has the type of "foo", however before the left-shift can be conducted it first must go through "integer promotion;" section 6.3.1.1.2 of the C99 standard specifies: "if an int can represent all values of the original type, the value is converted to an int."

这使你的code的非隐版（带额外的括号）如下：

This makes the non-implicit version of your code (with extra parentheses) as follows:

uint16_t foo;
foo = ((int)foo) << 1;

鉴于警告你是32位或64位整数的系统上（或任何比16大，真的）

，你确实推搡一个更大的值到一个较小的一个。

Given the warning you are on a system with either 32 or 64 bit ints (or anything bigger than 16, really), you are indeed shoving a bigger value into a smaller one.

解决此问题的方法是要明确你的演员，像这样：

One way around this is to be explicit with your casts like so:

uint16_t foo;
foo = (uint16_t)(foo << 1);

但是，这意味着，没有，你不能使用较短的按位移位赋值运算符。

But that means that no, you can not use the shorter bitwise shift assignment operator.

如果你真的这样做了一堆，考虑一个辅助功能，使您的code清晰，完全编译。

If you really are doing this a bunch, consider making a helper function that makes your code clear and compiles cleanly.

void LS(uint16_t &value, int shift) {  // LS means LeftShift
  *value = (uint16_t)(*value << shift);
}

LS(&foo, 1);

TL; DR：不，你不能使用短期操作者，避免在同一时间的警告

TL;DR: No, you can't use the short operator and avoid that warning at the same time.

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