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

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\n", mul);
	printf("0X%0X\n",mul);
	printf("signed: %d\n", m);
	printf("0X%0X\n",m);
	*/

	printf("%d\n",tmulOK(a,b));
	printf("%d\n",tmulOK2(a,b));
	printf("unsigned short: %d\n",tmulOK2(d,e));
	printf("%d\n",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\n",m);
	printf("l = 0X%0X\n",l);
	return (m & ~l) == 0;
}

/*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("\ntmulOK3()\n");
	printf("m =  0X%I64d\n",m);
	
	return m == (unsigned)m;
}