括号可以用C更改位运算的操作数的结果类型？ [英] Can parentheses in C change the result type of operands of a bitwise operation?

问题描述

我已经通过一个静态分析工具喂以下code：

I have fed the following code through a static analysis tool:

u1 = (u1 ^ u2); // OK

u1 = (u1 ^ u2) & u3;  // NOT OK

u1 = (u1 ^ u2) & 10; // NOT OK

u1 = (u1 ^ u2) & 10U; // NOT OK

u1 = (unsigned char)(u1 ^ u2) & 10U; // OK

u1 = (unsigned char)(u1 ^ u2) & u3;  // OK

OK是指静态分析工具，没有抱怨。
不OK指的是静态分析工具，并抱怨 - 声称按位操作的某些操作数不是一个无符号整数

"OK" means the static analysis tool did not complain. "NOT OK" means the static analysis tool did complain -- claiming that some operand of a bitwise operation is not an unsigned integer.

这最后2行中的结果表明，该括号是造成任

The results from the last 2 lines show that the parentheses are causing either

一个。要签署一个实际的类型转换

a. an actual type conversion to signed

乙。一些静态分析工具，认为是一种类型转换为签订

b. something that the static analysis tool thinks is a type conversion to signed

我会询问静态分析工具的开发（B）。

I will ask the static analysis tool developer about (b).

但我之前，我想知道，如果可能的C语言是众所周知的做（一）？

But before I do, I would like to know if perhaps the C language is known to do (a)?

推荐答案

用C没有下文 INT 完成：如增加了两个无符号的字符时，甚至加入前，操作数转换为 INT 根据默认的促销活动。

Nothing in C is done below int: eg when adding two unsigned chars, even before the addition, the operands are converted to int according to the default promotions.

unsigned char u1, u2, u3;
u1 = 0;
u2 = 42;
u3 = u1 + u2;

在最后一行，第一个 U1 和 U2 转换为 INT ，那么 + 运算符应用于获得 INT 值，然后该值被转换回到 unsigned char型（当然编译器可以使用快捷键！）

In the last line, first u1 and u2 are converted to int, then the + operator is applied to obtain a int value and then that value is converted back to unsigned char (of course the compiler can use shortcuts!)

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