Re: [問題]有關 round() 的問題

[Schottky (Schottky)]

※ 引述《watson1988 (watson)》之銘言：
: 四捨五入還有其他更簡單的寫法嗎!?

寫了個程式測試...
我們希望標準 ROUND() 的行為應該是:
ROUND( 1.49) =  1.000000
ROUND( 1.5)  =  2.000000
ROUND( 1.51) =  2.000000
ROUND(-1.49) = -1.000000
ROUND(-1.5)  = -2.000000
ROUND(-1.51) = -2.000000

執行結果是 ROUND3 和 ROUND5 這兩種作法才是正解...
我說的 #define ROUND2 floor((x)+0.5) 會造成 ROUND2(-1.5) == -1 這樣也不對,
此外順便炫耀一下華麗的 macro ... :D

// URL: http://codepad.org/8gI4wz2b
#include <stdio.h>
#include <math.h>

#define ROUND1(x) ((x)-(int)(x))>=0.5?ceil((x)):floor((x))
#define ROUND2(x) floor((x)+0.5)
#define ROUND3(x) (double)(int)((x) + 0.5 - (double)((x) < 0))
#define ROUND4(x) ((x)-floor(x))>=0.5?ceil((x)):floor((x))
#define ROUND5(x) ((x)>0?floor((x)+0.5):ceil((x)-0.5))

#define __T(x) #x
#define _T(x) __T(x)

#define DEF(y) \
    printf("#define " #y "(x) " _T(y(x)) "\n"); \
    printf(#y "( 1.49) = % f\n", y(1.49));  \
    printf(#y "( 1.5)  = % f\n", y(1.5));   \
    printf(#y "( 1.51) = % f\n", y(1.51));  \
    printf(#y "(-1.49) = % f\n", y(-1.49)); \
    printf(#y "(-1.5)  = % f\n", y(-1.5));  \
    printf(#y "(-1.51) = % f\n", y(-1.51)); \
    printf("\n");

int main(void)
{
    DEF(ROUND1);
    DEF(ROUND2);
    DEF(ROUND3);
    DEF(ROUND4);
    DEF(ROUND5);

    return 0;
}

// 執行結果:
#define ROUND1(x) ((x)-(int)(x))>=0.5?ceil((x)):floor((x))
ROUND1( 1.49) =  1.000000
ROUND1( 1.5)  =  2.000000
ROUND1( 1.51) =  2.000000
ROUND1(-1.49) = -2.000000
ROUND1(-1.5)  = -2.000000
ROUND1(-1.51) = -2.000000

#define ROUND2(x) floor((x)+0.5)
ROUND2( 1.49) =  1.000000
ROUND2( 1.5)  =  2.000000
ROUND2( 1.51) =  2.000000
ROUND2(-1.49) = -1.000000
ROUND2(-1.5)  = -1.000000
ROUND2(-1.51) = -2.000000

#define ROUND3(x) (double)(int)((x) + 0.5 - (double)((x) < 0))
ROUND3( 1.49) =  1.000000
ROUND3( 1.5)  =  2.000000
ROUND3( 1.51) =  2.000000
ROUND3(-1.49) = -1.000000
ROUND3(-1.5)  = -2.000000
ROUND3(-1.51) = -2.000000

#define ROUND4(x) ((x)-floor(x))>=0.5?ceil((x)):floor((x))
ROUND4( 1.49) =  1.000000
ROUND4( 1.5)  =  2.000000
ROUND4( 1.51) =  2.000000
ROUND4(-1.49) = -1.000000
ROUND4(-1.5)  = -1.000000
ROUND4(-1.51) = -2.000000

#define ROUND5(x) ((x)>0?floor((x)+0.5):ceil((x)-0.5))
ROUND5( 1.49) =  1.000000
ROUND5( 1.5)  =  2.000000
ROUND5( 1.51) =  2.000000
ROUND5(-1.49) = -1.000000
ROUND5(-1.5)  = -2.000000
ROUND5(-1.51) = -2.000000
※ 編輯: Schottky        來自: 118.167.25.242       (10/03 20:08)
→ Feis:印象中 g++ 好像用 trunc() 去實作 round() (很不確定) XD
推 EdisonX:這問題小弟有探討過，這類型前提都是要假設轉型後沒溢位。
→ EdisonX:http://ppt.cc/CFLR 參考
推 damody:一下就五個 round 是哪招= =
→ Schottky:對喔,我忘了float不能隨便轉int... :p
→ Feis:+-0.5 也是個很可能爆炸的問題 @@
→ Schottky:+-0.5 有什麼問題?
推 watson1988:謝謝專業的s大