1<< 31产生错误，“<<'的结果，表达式未定义“

问题描述

如果我在32位平台上将值声明为int类型并执行以下操作：

  int32_t mask; 
 mask = 1<< 31：//产生2147483648（或0x80000000）
  

有人可以帮助我理解上述行为什么会产生警告：

 '<<'表达式的结果未定义
  
 
 
 
 
 int32_t ，它从-2 到（2  -1）。这是C11和C ++ 11中未定义的行为，并且在C ++ 14中实现定义。
 
 
 
 C11§6.5.7/ p4（引用N1570） / p> 
 
 
 N3337中的C ++ 11规则§5.8[expr.shift] / p2几乎完全相同。因为2 不可表示，所以行为是未定义的。
 
 
 
 C ++ 14§5.8[expr.shift] / p2 N3936;另见）： 
 As 2 位int，行为被定义，结果是2 转换为 int32_t ;这个转换是根据§4.7[conv.integral] / p3的实现定义。在一个使用二进制补码的典型系统中，你会得到-2 。
 
If I have value declared as of type int on a 32 bit platform and perform the following:
int32_t mask;
mask = 1 << 31: // produces 2147483648 (or 0x80000000)
Can someone help me understand why the above line would produce the warning:
The result of the '<<' expression is undefined
 解决方案 
2 isn't representable in a int32_t, which goes from -2 to (2-1). This is undefined behavior in C11 and C++11, and implementation-defined in C++14.
C11 §6.5.7/p4 (quoting N1570):
The C++11 rule in N3337 §5.8 [expr.shift]/p2  is pretty much identical. Since 2 isn't representable, the behavior is undefined.
C++14  §5.8 [expr.shift]/p2 (quoting N3936; see also CWG issue 1457):
As 2 is representable in an unsigned 32-bit int, the behavior is defined and the result is 2 converted to int32_t; this conversion is implementation-defined per §4.7 [conv.integral]/p3. In a typical system using two's complement you'd get -2.
                        
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