向右移位并有符号整数 [英] Right shift and signed integer
问题描述
在我的编译器，下面的伪code（值二进制替换）：
On my compiler, the following psuedo code (values replaced with binary):
sint32 word = (10000000 00000000 00000000 00000000);
word >>= 16;
生成一个字与位域，如下所示：
produces a word with a bitfield that looks like this:
(11111111 11111111 10000000 00000000)
我的问题，是我可以依靠此行为对所有平台和C ++编译器？
My question, is can I rely on this behaviour for all platforms and C++ compilers?
推荐答案
从以下链接：
<一href="https://www.securecoding.cert.org/confluence/display/sec$c$c/INT34-C.+Do+not+shift+an+ex$p$pssion+by+a+negative+number+of+bits+or+by+greater+than+or+equal+to+the+number+of+bits+that+exist+in+the+operand"相对=nofollow> INT34-C。不要位的负数或大于或等于位中存在的操作数的数目移位的当然pression
From the following link:
INT34-C. Do not shift an expression by a negative number of bits or by greater than or equal to the number of bits that exist in the operand
不合规code为例（右Shift）
结果 E1＆GT;＆GT; E2 是 E1 右移 E2 位的位置。如果 E1 有一个无符号的类型，或者 E1 有符号类型和非负值，则结果的值E1 / 2 E2 的商的整数部分。如果 E1 有签署类型和负值，得到的值是实现定义的，可以是一个算术（签字）转变：
或逻辑（符号）转变：
这不符合code例如无法测试右操作数是否大于或等于促进左操作数的宽度，使不确定的行为。
Noncompliant Code Example (Right Shift)
The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2E2. If E1 has a signed type and a negative value, the resulting value is implementation defined and can be either an arithmetic (signed) shift:
or a logical (unsigned) shift:
This noncompliant code example fails to test whether the right operand is greater than or equal to the width of the promoted left operand, allowing undefined behavior.
unsigned int ui1;
unsigned int ui2;
unsigned int uresult;
/* Initialize ui1 and ui2 */
uresult = ui1 >> ui2;
作出假设右移是否实现为算术（签字）移位或逻辑（无符号）移也可能导致安全漏洞。见建议<一href="https://www.securecoding.cert.org/confluence/display/sec$c$c/INT13-C.+Use+bitwise+operators+only+on+unsigned+operands"相对=nofollow> INT13-C。使用位运算符只能在无符号运算的。
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