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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