问题描述
我想完成需要我写三种功能为二进制算术的分配。 BADD（）是对我提供的，所以我用它来帮助编写bsub（）和bmult（）函数。我无法理解我应该怎么然而执行BDIV（）函数。我知道我需要通过使用右移和我bsubb（）函数位迭代，但我不知道如何实现它。下面是我到目前为止写的功能。让我知道你是否注意到我在写他们犯任何错误（即bsub（）和bmult（））。谢谢。
/ **此功能增加了使用位运算符的两个参数。你的
*实现不应该使用算术运算符除循环
* 控制。整数是32位长。这个函数打印一条消息
*说：发生溢出\\ N如果发生二进制补码溢出
*在加入过程。的总和作为返回值
*该函数。
* /
INT BADD（INT X，int y）对{INT I;焦总和;
焦炭car_in = 0;
焦炭car_out;
烧焦A，B;unsigned int类型面膜= 00000001;
INT结果为0;对于（i = 0; I＆LT; 32;我++）{ 一个=（X安培;掩模）！= 0;
B =（Y＆安培;面罩）！= 0;
car_out = car_in＆安培; （A | B）|一个和b;
总和= A ^ B ^ car_in; 如果（和）{
结果| =口罩;
} 如果（我！= 31）{
car_in = car_out;
}其他{
如果（car_in！= car_out）{
的printf（溢出发生\\ n）;
}
} 面膜＆LT;＆LT; = 1;
} 返回结果;
}//通过查找compliemnt subracts两个整数
Y，加1，并使用BADD的//（）函数
//添加-y和x
INT bsub（INT X，int y）对{返回BADD（X，BADD（〜Y，1））;
}
//添加X总额为然而，许多ÿ
INT bmult（INT X，int y）对{INT总;
INT I;
对于（i = 0; I＆LT; = Y;我++）
{
总= BADD（总数，x）的
}
总回报;
}//我的评论
unsigned int类型BDIV（unsigned int类型红利，无符号整型除数）{//给我写信
返回0;
}
这里就不多说了，它在只是一些基本的数学基础-2：
unsigned int类型bmult（unsigned int类型的x，unsigned int类型Y）
{
INT总= 0;
INT I; / *如果第i位为非零，添加的x到总* /
/ *多总2每一步* /
对于（I = 32; I＆GT; = 0;我 - ）
{
总＆LT;＆LT; = 1;
如果（（Y及（1下;＆所述; i））的＆GT; I标记）
{
总= BADD（总的，x）的;
}
} 总回报;
}unsigned int类型BDIV（unsigned int类型红利，无符号整型除数）
{
INT I，商= 0，余数= 0; 如果（除数== 0）{printf的（分区零\\ n）;返回0; } 对于（I = 31; I＆GT; = 0;我 - ）
{
商数＆所述;＆下; = 1;
其余与所述;＆下; = 1;
余| =（分红及（1 LT;＆LT; I））＆GT;＆GT;一世; 如果（余＆GT; =除数）
{
余数= bsub（余，除数）;
商| = 1;
}
} 返回商;
}
这两篇文章都足以code这些样本：和。
I am trying to complete an assignment that requires me to write three functions for binary arithmetic. badd() was provided for me, so I used it to help write the bsub() and bmult() functions. I am having trouble understanding how I should perform the bdiv() function, however. I know I need to iterate through the bits using a right shift and my bsubb() function, but I don't know how to implement it. Below are the functions that I have written so far. Let me know if you notice any mistakes that I made in writing them(meaning bsub() and bmult()). Thanks.
/** This function adds the two arguments using bitwise operators. Your
* implementation should not use arithmetic operators except for loop
* control. Integers are 32 bits long. This function prints a message
* saying "Overflow occurred\n" if a two's complement overflow occurs
* during the addition process. The sum is returned as the value of
* the function.
*/
int badd(int x,int y){
int i;
char sum;
char car_in=0;
char car_out;
char a,b;
unsigned int mask=0x00000001;
int result=0;
for(i=0;i<32;i++){
a=(x&mask)!=0;
b=(y&mask)!=0;
car_out=car_in & (a|b) |a&b;
sum=a^b^car_in;
if(sum) {
result|=mask;
}
if(i!=31) {
car_in=car_out;
} else {
if(car_in!=car_out) {
printf("Overflow occurred\n");
}
}
mask<<=1;
}
return result;
}
// subracts two integers by finding the compliemnt
// of "y", adding 1, and using the badd() function
// to add "-y" and "x"
int bsub(int x, int y){
return badd(x, badd(~y, 1));
}
//add x to total for however many y
int bmult(int x,int y){
int total;
int i;
for(i=0; i < = y; i++)
{
total = badd(total,x)
}
return total;
}
// comment me
unsigned int bdiv(unsigned int dividend, unsigned int divisor){
// write me
return 0;
}
Not much to say here, it's just some basic math in base-2:
unsigned int bmult(unsigned int x, unsigned int y)
{
int total = 0;
int i;
/* if the i-th bit is non-zero, add 'x' to total */
/* Multiple total by 2 each step */
for(i = 32 ; i >= 0 ; i--)
{
total <<= 1;
if( (y & (1 << i)) >> i )
{
total = badd(total, x);
}
}
return total;
}
unsigned int bdiv(unsigned int dividend, unsigned int divisor)
{
int i, quotient = 0, remainder = 0;
if(divisor == 0) { printf("div by zero\n"); return 0; }
for(i = 31 ; i >= 0 ; i--)
{
quotient <<= 1;
remainder <<= 1;
remainder |= (dividend & (1 << i)) >> i;
if(remainder >= divisor)
{
remainder = bsub(remainder, divisor);
quotient |= 1;
}
}
return quotient;
}
These two articles are enough to code these samples: Div and Mul.
这篇关于执行位无师算术运算符的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！