问题描述
我从下面的代码中得到错误的o / p：
#include< stdio.h>
#include< float.h>
#include< limits.h>
main（）
{
双倍val，val1;
val = 123456789012.123456;
val1 = 123456789012.123456;
val2 = val * val1;
printf（"％f\ n"，val2）;
}
我得到的结果是15241578753183967100000.000000而不是正确的值。
我在带有ansi c编译器的HP-UX 11上运行。
任何人都可以告诉我为什么我会得到这个结果。
Hi,
I am getting incorrect o/p from the code below:
#include<stdio.h>
#include<float.h>
#include<limits.h>
main()
{
double val,val1;
val=123456789012.123456;
val1=123456789012.123456;
val2=val*val1;
printf("%f\n",val2);
}
The result I get is 15241578753183967100000.000000 instead of the correct value.
I am running on HP-UX 11 with ansi c compiler.
Can anyone let me know why I am getting this result.
推荐答案
它看起来足够接近我。你有什么回答？
- glen
It looks close enough for me. What answer did you expect?
-- glen
看起来val2默认为int类型，而你正试图
打印出来就像双人一样。
- James
It looks like val2 is defaulting to an int type, and you are trying to
print it out like a double.
-- James
这似乎是你的机器能够出现的最佳近似值
- 浮点变量只有有限的精度。并且18
位数看起来并不太糟糕。欢迎来到美好的世界
浮点数学;-)因为你已经包含了< float.h>
检查编译器承诺了多少位数你打印出来的价格是
DBL_DIG的值。
问候，Jens
-
_ _____ _____
| || _ _ || _ _ |
_ | | | | | |
| | _ | | | | | |
\ ___ / ens | _ | homs | _ | oerring
That seems to the best approximation your machine is able to come up
with - floating point variables have only a finite precision. And 18
digits doesn''t look too bad. Welcome to the wonderful world of
floating point math ;-) And since you''re already including <float.h>
check how many digits the compiler is promising you by printing out
the value of DBL_DIG.
Regards, Jens
--
_ _____ _____
| ||_ _||_ _| Je***********@physik.fu-berlin.de
_ | | | | | |
| |_| | | | | | http://www.physik.fu-berlin.de/~toerring
\___/ens|_|homs|_|oerring
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