整数加法考虑CARRY [英] integer addition taking CARRY into account
问题描述
我使用的编译器不支持long int（32位）
所以我使用2 int'来做工作
我的问题
int a [2];
int b [2];
int c [2];
我想
c [0] = a [0] + b [0]; ----->第一步
c [1] = a [1] + b [1] +从第一步开始
如何从第一步获得进位C编程？
任何提示？
感谢您的时间，
archilles
" ar **************** @ hotmail.com" < AR **************** @ hotmail.com>写道：
我使用的编译器不支持long int（32位）
所以我使用2 int'来做这项工作
我的问题
int a [2];
int b [2];
int c [2];
我想要
c [0] = a [0] + b [0]; ----->第一步
c [1] = a [1] + b [1] +从第一步开始
如果无符号整数可以接受（替换`int ''通过'unsigned''
以上），然后我相信以下内容将起作用：
c [0] = a [0] + b [0];
c [1] = a [1] + b [1] +（c [0]< a [0] || c [0]< b [0]）;
如果错误，我肯定有人会纠正我。
-
int main（void）{char p [] =" ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz。 \
）; while（* q）{i + = strchr（p，* q ++） - p; if（i> =（int）sizeof p）i- = sizeof p-1; putchar（p [i] \
）;}返回0;}
Ben Pfaff写道：" ar ***** ***********@hotmail.com" < ar **************** @ hotmail.com>
写道：我使用的编译器不支持long int（32位）
所以我使用2 int'来做工作
我的问题
int a [2];
int b [2];
int c [2];
我想
c [0] = A [0] + b [0]; ----->第一步
c [1] = a [1] + b [1] +从第一步开始
如果无符号整数是可接受的（用`替换`int''上面没有签名''
，我认为以下内容可行：
c [0] = a [0] + b [0];
c [1] = a [1 ] + b [1] +（c [0]< a [0] || c [0]< b [0]）;
或。 ..
c [0] = a [0] + b [0];
c [1] = a [1] + b [1] + （a [0]> -1u - b [0]）;
-
彼得
在文章< 87 ************ @ benpfaff.org>，
Ben Pfaff< bl*@cs.stanford。 EDU>写道：
如果无符号整数是可以接受的（将'int''替换为'unsigned''
，那么我相信以下内容将会起作用：
c [0] = a [0] + b [0];
c [1] = a [1] + b [1] +（c [0]< a [0] || c [0 ]< b [0]）;
携带的表达式不必要地复杂化。如果在第一次添加时出现溢出
，则c [0]将小于* * a [0]和
b [0]，因此您可以将其与其中一个：
c [1] = a [1] + b [1] +（c [0]< a [0]）;
- Richard
Hi,
I am using a compiler that does not support long int (32 bits)
so I am using 2 int''s to do the work
my problem
int a[2];
int b[2];
int c[2];
I want to
c[0]=a[0]+b[0]; -----> Step one
c[1]=a[1]+b[1]+carry from Step one
how do I get the carry from step one in C programming ?
Any hints ?
thanks for your time,
archilles
"ar****************@hotmail.com" <ar****************@hotmail.com> writes:
I am using a compiler that does not support long int (32 bits)
so I am using 2 int''s to do the work
my problem
int a[2];
int b[2];
int c[2];
I want to
c[0]=a[0]+b[0]; -----> Step one
c[1]=a[1]+b[1]+carry from Step one
If unsigned integers are acceptable (replace `int'' by `unsigned''
above), then I believe that the following will work:
c[0] = a[0] + b[0];
c[1] = a[1] + b[1] + (c[0] < a[0] || c[0] < b[0]);
I am sure someone will correct me if it is wrong.
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}
Ben Pfaff wrote:"ar****************@hotmail.com" <ar****************@hotmail.com>
writes:I am using a compiler that does not support long int (32 bits)
so I am using 2 int''s to do the work
my problem
int a[2];
int b[2];
int c[2];
I want to
c[0]=a[0]+b[0]; -----> Step one
c[1]=a[1]+b[1]+carry from Step one
If unsigned integers are acceptable (replace `int'' by `unsigned''
above), then I believe that the following will work:
c[0] = a[0] + b[0];
c[1] = a[1] + b[1] + (c[0] < a[0] || c[0] < b[0]);
Or...
c[0] = a[0] + b[0];
c[1] = a[1] + b[1] + (a[0] > -1u - b[0]);
--
Peter
In article <87************@benpfaff.org>,
Ben Pfaff <bl*@cs.stanford.edu> wrote:
If unsigned integers are acceptable (replace `int'' by `unsigned''
above), then I believe that the following will work:
c[0] = a[0] + b[0];
c[1] = a[1] + b[1] + (c[0] < a[0] || c[0] < b[0]);
The expression for carry is unnecessarily complicated. If overflow
occurs in the first addition, c[0] will be less than *both* a[0] and
b[0], so you can compare it with either one:
c[1] = a[1] + b[1] + (c[0] < a[0]);
-- Richard
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