判断C语言中int 与 unsigned 乘法是否会溢出

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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\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;}


	
				
		
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