右移和有符号整数 [英] Right shift and signed integer

问题描述

在我的编译器上，下面的伪代码（用二进制替换的值）：

  sint32 word =（10000000 00000000 00000000 00000000 ）; 
 word>> = 16; 
  

产生一个字像这样：

 （11111111 11111111 10000000 00000000）
  

我的问题是，我可以依赖于所有平台和C ++编译器的这种行为。

解决方案

从以下链接：

。

On my compiler, the following psuedo code (values replaced with binary):

sint32 word = (10000000 00000000 00000000 00000000);
word >>= 16;

produces a word with a bitfield that looks like this:

(11111111 11111111 10000000 00000000)

My question, is can I rely on this behaviour for all platforms and C++ compilers?

解决方案

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

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 / 2^E2. 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;

Making assumptions about whether a right shift is implemented as an arithmetic (signed) shift or a logical (unsigned) shift can also lead to vulnerabilities. See recommendation INT13-C. Use bitwise operators only on unsigned operands.

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