有没有四舍无入的函数? - 未名空间MITBBS历史存档

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有没有四舍无入的函数?

有没有四舍无入的函数?# Unix - 噫吁兮，危乎高哉

d*r2000-10-23 07:10

1 楼

我知道int是取整,有没有类似的取整函数但是算法是四舍五入的?

y*y2000-10-23 07:10

2 楼

why not write by yourself? it's so simple.

【在 d******r 的大作中提到】

: 我知道int是取整,有没有类似的取整函数但是算法是四舍五入的?

l*l2000-10-23 07:10

3 楼

int(x+0.5)

【在 y***y 的大作中提到】

: why not write by yourself? it's so simple.

y*y2000-10-23 07:10

4 楼

what if x is equal to interger.

【在 l*l 的大作中提到】

: int(x+0.5)

bz2000-10-23 07:10

5 楼

no, lib functions are all doing round off, but writing one is very
easy.

【在 d******r 的大作中提到】

: 我知道int是取整,有没有类似的取整函数但是算法是四舍五入的?

h*h2000-10-23 07:10

6 楼

take it easy, think about so many non-cs majors changed to
Computer Science.
It is always hard to get started.

【在 l*l 的大作中提到】

: int(x+0.5)

y*y2000-10-23 07:10

7 楼

He deleted his post because he know he made a mistake.
BUt it seems you still dont understand and self-feeling good.
What I mean is int(x+0.5) gives you x if x is an interger itself.
but 四舍五入 gives you x+1. so int(x+0.5) is a wrong answer. You need
to write if-else.

【在 h**h 的大作中提到】

: take it easy, think about so many non-cs majors changed to
: Computer Science.
: It is always hard to get started.

y*y2000-10-23 07:10

8 楼

faint...it seems I made the mistake....anyway,sometimes brain
stalk....:(

【在 y***y 的大作中提到】

: He deleted his post because he know he made a mistake.
: BUt it seems you still dont understand and self-feeling good.
: What I mean is int(x+0.5) gives you x if x is an interger itself.
: but 四舍五入 gives you x+1. so int(x+0.5) is a wrong answer. You need
: to write if-else.

y*y2000-10-23 07:10

9 楼

okay, I made a mistake, but the int(x+0.5) is still a wrong answer.
hohohoho~

【在 y***y 的大作中提到】

: faint...it seems I made the mistake....anyway,sometimes brain
: stalk....:(

y*y2000-10-23 07:10

10 楼

try some negative digits.
you still need to write if-else.

【在 y***y 的大作中提到】

: okay, I made a mistake, but the int(x+0.5) is still a wrong answer.
: hohohoho~

y*y2000-10-23 07:10

11 楼

正解:
#!/usr/local/perl
$x=;
if($x>=0){
print int($x+0.5);
}
else{
print int($x-0.5);}
win!! hohohoho~

【在 y***y 的大作中提到】

: try some negative digits.
: you still need to write if-else.

p*h2000-10-23 07:10

12 楼

how about this:
int i;
double d;
/* d = whatever number. */
i = (int)floor(d + 0.5);

【在 y***y 的大作中提到】

: 正解:
: #!/usr/local/perl
: $x=;
: if($x>=0){
: print int($x+0.5);
: }
: else{
: print int($x-0.5);}
: win!! hohohoho~

c*t2000-10-23 07:10

13 楼

Sigh, you are the confused one :)
For the negative numbers, there are two rounding schemes.

【在 y***y 的大作中提到】

: 正解:
: #!/usr/local/perl
: $x=;
: if($x>=0){
: print int($x+0.5);
: }
: else{
: print int($x-0.5);}
: win!! hohohoho~

m*e2000-10-23 07:10

14 楼

Mod maths treats it the same no matter you are + or -. Always the largest
int that is no larger than the number. This is consistent.

【在 c*****t 的大作中提到】

: Sigh, you are the confused one :)
: For the negative numbers, there are two rounding schemes.

l*l2000-10-23 07:10

15 楼

you are smart, but the code is too long, even it is sample.
I write one line and solve it, I guess it is the best code any one can
get.
print int(($x+abs($x))/2+0.5)+int(($x-abs($x))/2-0.5);

【在 y***y 的大作中提到】

: try some negative digits.
: you still need to write if-else.

b*d2000-10-23 07:10

16 楼

Best? How do you evaluate if a program is good or poor?
Do you think such code is in good style?

【在 l*l 的大作中提到】

: you are smart, but the code is too long, even it is sample.
: I write one line and solve it, I guess it is the best code any one can
: get.
: print int(($x+abs($x))/2+0.5)+int(($x-abs($x))/2-0.5);