比较尾数 [英] comparing mantissa
问题描述
嗨！
是否有更好/更快的方式来比较mantissas到实数然后
以下代码？
#include< stdio.h>
#include< stdlib.h>
int main（void）{
浮动a，b;
int test;
a = 1.4234;
b = 3.34;
test =（（a-（int）a）>（b-（int）b））;
printf（" a =％fb =％f \ n mantisa更大在％d，a，b，测试中）;
返回EXIT_SUCCESS;
}
谢谢提前你
Hi!
is there a better/faster way to compare mantissas of to real number then
in following code?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
float a,b;
int test;
a=1.4234;
b=3.34;
test=((a-(int)a)>(b-(int)b));
printf("a=%f b=%f\n mantisa is bigger in a %d", a,b,test);
return EXIT_SUCCESS;
}
Thank you in advance
推荐答案
Carramba写道：
Carramba wrote:
嗨！
有更好/更快的方法来比较mantissas到实数，然后
在下面的代码中？
#include< stdio.h>
#include< stdlib.h>
int main（无效）{
浮动a，b;
int test;
a = 1.4234;
b = 3。 34;
test =（（a-（int）a）>（b-（int）b））;
printf（" a =％fb = ％f \ n mantisa在％d，a，b，test中更大;
返回EXIT_SUCCESS;
}
Hi!
is there a better/faster way to compare mantissas of to real number then
in following code?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
float a,b;
int test;
a=1.4234;
b=3.34;
test=((a-(int)a)>(b-(int)b));
printf("a=%f b=%f\n mantisa is bigger in a %d", a,b,test);
return EXIT_SUCCESS;
}
你的意思是< math.hmodf（）？在数值分析的意义上，它看起来并不像你的意思是
。
Do you mean <math.hmodf()? It doesn''t look like you mean mantissa in
the numerical analysis sense.
好吧，如果你不关心exponenet只关心分数的大小
除了特殊情况（NaN，+ / - 无穷大）之外你可以
得到指向int和point的指针它可以记忆数字的内存位置。
将指向int（实际上是浮点数）的数据移到指向左边的数据$
但是您的解决方案：仍然保留数字的符号部分。那么，
-.222< .111，按照你的规范-.222应该大于.111。
---马修希克斯
Well, if you don''t care about the exponenet and only care about the size
of the fraction, in all but the special cases (NaN, +/- infinity) you can
get a pointer to int and point it to the memory location that sores the number.
Shift the data that the pointer to int (really a float) points to left by
nine bits. Do the same for the other number and then compare the results.
But your "solution: still maintains the sign portion of the number. So,
-.222 < .111, by your specification -.222 should be greater than .111.
---Matthew Hicks
你好！
是否有更好/更快的方式来比较mantissas到实数
然后
以下代码？
#include< stdio.h>
#include< stdlib.h>
int main（void）{
浮动a ，b;
int test;
a = 1.4234;
b = 3.34;
test =（（a- （int）a）>（b-（int）b））;
printf（a =％fb =％f \ n mantisa在％d中更大，a， b，测试）;
返回EXIT_SUCCESS;
}
提前谢谢
Hi!
is there a better/faster way to compare mantissas of to real number
then
in following code?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
float a,b;
int test;
a=1.4234;
b=3.34;
test=((a-(int)a)>(b-(int)b));
printf("a=%f b=%f\n mantisa is bigger in a %d", a,b,test);
return EXIT_SUCCESS;
}
Thank you in advance
" Carramba" < ca ****** @ example.comwrote in message
news：f7 ********** @ aioe.org ...
"Carramba" <ca******@example.comwrote in message
news:f7**********@aioe.org...
嗨！
有更好/更快的方法来比较mantissas到实数，然后
在下面的代码中？
#include< stdio.h>
#include< stdlib.h>
int main（void）{
浮动a，b;
int test;
a = 1.4234;
b = 3.34;
test =（（a-（int）a）>（b-（int）b））;
printf（" a =％fb =％f \ n mantisa在％d，a，b，测试中更大;
返回EXIT_SUCCESS;
}
提前谢谢
Hi!
is there a better/faster way to compare mantissas of to real number then
in following code?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
float a,b;
int test;
a=1.4234;
b=3.34;
test=((a-(int)a)>(b-(int)b));
printf("a=%f b=%f\n mantisa is bigger in a %d", a,b,test);
return EXIT_SUCCESS;
}
Thank you in advance
是的，还有更好的方法，因为上面的内容将无法给出正确的
回答负数。
-
Fred L. Kleinschmidt
波音助理技术研究员
Aero Stability and控制计算
Yes, there is a better way, since the above will fail to give the correct
answer for negative numbers.
--
Fred L. Kleinschmidt
Boeing Associate Technical Fellow
Aero Stability and Controls Computing
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