有符号字符的乘法问题 [英] Problem with multiplication of signed chars

问题描述

讨论这个的答案我发现，下面的代码打印 -1 和 1 。为什么？在我看来，它应该在乘法期间打印两个 1 的溢出溢出。

  signed char c1 = numeric_limits< signed char> :: min（）; 
 signed char c2 = -1; 
 cout< c1 * c2 / c1 < endl; 
 signed char result = c1 * c2; 
 cout< result / c1<< endl; 
  

解决方案

c1 可能具有像 -128 的值，例如。在乘法中，整数促销将导致 c1 和 c2 转换为 int

c1

int 与值 128 因此 c1 * c2 / c1 将成为 int ，值为 -1 。

-1 对于第一个输出看起来正确。

对于第二个版本， c1 * c2 的结果将不适合 signed char ，并且将转换为实现定义的结果， -128 而不是 128 。

Discussing this answer I find out that the code bellow prints -1 and 1 in visual studio. Why? In my opinion it should print two 1s despit overflow during multiplication.

signed char c1 = numeric_limits<signed char>::min();
signed char c2 = -1;
cout << c1 * c2 / c1 << endl;
signed char result = c1 * c2;
cout << result / c1 << endl;

解决方案

c1 might have a value like -128, say. In the multiplication, integer promotions will cause both c1 and c2 to be converted to type int before the operation is performed.

c1 * c2 is then going to be a int with value 128 so c1 * c2 / c1 is going to be an int with value -1.

-1 for the first output looks correct to me.

For the second version, typically the assignment of the result of c1 * c2 won't fit into a signed char and will convert to an implementation-defined result, perhaps -128 instead of 128.

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