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判斷C語言中int 與 unsigned 乘法是否會溢出

日期:2017/3/1 10:00:33   编辑:Linux編程

在C語言中,int 與 unsigned 乘法被定義為產生w(w為機器字長)位的值。如果乘積超過w位,所產生乘積的高位將被捨棄。

下面這段代碼用來判斷整數乘法會不會溢出:

/*練習題2.36*/
/*開發環境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));
}

/*判斷兩整數相乘是否溢出,不溢出則返回1*/
int tmulOK(int x, int y){
int p = x * y;
return !x || p/x == y;
}

/*判斷兩整數相乘是否溢出,不溢出則返回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;
}
<PRE class=cpp name="code">/*判斷兩整數相乘是否溢出,不溢出則返回1*/
int tmulOK3(unsigned x, unsigned y){
_int64 m = (_int64)x * y; /*_int64(也可寫為__int64)為64位整數。此處的(_int64)強制類型轉換相當重要,如果不加此強制類型轉換則
x*y就會按照32位乘法進行運算,這樣乘積中高出的32位的更高位將被捨棄。*/
&nbsp; printf("\ntmulOK3()\n");
printf("m = 0X%I64d\n",m);

return m == (unsigned)m;
}

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