判断C语言中int 与 unsigned 乘法是否会溢出

在C语言中，int 与 unsigned 乘法被定义为产生w（w为机器字长）位的值。如果乘积超过w位，所产生乘积的高位将被舍弃。下面这段代码用来判断整数乘法会不会溢出：/*练习题2.36*/
/*开发环境VC++ 6.0*/
#include<stdio.h>void main（）{
unsigned x = 4294967295;
unsigned y = 8;
unsigned mul = x * y; int a = 2147483647;
int b = 8;
int m = a * b; unsigned short d = 65535;
unsigned short e = 1; /*
printf（"unsigned: %u ", mul）;
printf（"0X%0X ",mul）;
printf（"signed: %d ", m）;
printf（"0X%0X ",m）;
*/ printf（"%d ",tmulOK（a,b））;
printf（"%d ",tmulOK2（a,b））;
printf（"unsigned short: %d ",tmulOK2（d,e））;
printf（"%d ",tmulOK3（x,y））;
}/*判断两整数相乘是否溢出，不溢出则返回1*/
int tmulOK（int x, int y）{
int p = x * y;
return ！x || p/x == y;
}/*判断两整数相乘是否溢出，不溢出则返回1*/
int tmulOK2（unsigned short x, unsigned short y）{
int m = x * y;

unsigned short i = ~0;
int l = i; printf（"m = 0X%0X ",m）;
printf（"l = 0X%0X ",l）;
return （m & ~l） == 0;
}
<PRE class=cpp name="code">/*判断两整数相乘是否溢出，不溢出则返回1*/
int tmulOK3（unsigned x, unsigned y）{
_int64 m = （_int64）x * y; /*_int64（也可写为__int64）为64位整数。此处的（_int64）强制类型转换相当重要，如果不加此强制类型转换则
x*y就会按照32位乘法进行运算，这样乘积中高出的32位的更高位将被舍弃。*/
  printf（" tmulOK3（） "）;
printf（"m = 0X%I64d ",m）;

return m == （unsigned）m;
}