标题: 浮点数和16进制的相互转换工具 [打印本页] 作者: shane007 时间: 2010-2-8 11:32 标题: 浮点数和16进制的相互转换工具 在很多图片字库的游戏里用到了浮点数，这个转换工具也许会派上用处。) K6 N+ d3 J- F" x+ {
工具是我用C代码自己编译的。6 Q( x$ `5 p x) G
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原文3 z3 \4 J) @$ Y. a( _ http://www.cs.cornell.edu/~tomf/notes/cps104/floating.html/ z8 [) q3 h z- [2 k
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Helper Software 7 V$ k+ |5 d2 ?) c' ^6 K' dIf you're interested in investigating further, I include two programs for which I provide the C code that you can run to gain a greater understanding of how floating point works, and also to check your work on various assignments.) g& I( ]% n9 \; V1 z' E) ~
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Hex 2 Float + o% i/ t) X6 V5 {: n0 E4 f( A#include <stdio.h>: o3 a- V3 v; ~ u9 f' [
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int main()" @, {$ }, D& W
{ 8 e2 ^6 M0 B7 D) K( r0 d( Q float theFloat;, C: j% Z6 `- U1 W3 _
while (1) {) a" h7 r, w& q, ^3 G; B) x
scanf("%x", (int *) &theFloat);+ I4 q6 v. \( w3 y8 w
printf("0x%08X, %f\n", *(int *)&theFloat, theFloat);* S7 Y4 l9 \! y
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return 0;0 @8 P# _2 o2 C* e
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This program accepts as input a hexadecimal quantity and reads it as raw data into the variable "theFloat." The program then outputs the hexadecimal representation of the data in "theFloat" (repeating the input), and prints alongside it the floating point quantity that it represents. + e. U$ D! l2 h) t q E5 {# ] ; \4 a0 A6 T+ L7 b; S2 \; _5 U7 X" SI show here a sample run of the program. Notice the special case floating point quantities (0, infinity, and not a number). 8 P% h m' _: u$ n7 _' d; x* G2 @! \5 k+ O) j
For the denormalized but nonzero numbers, this program will display zero even though the number is not really zero. If you want to get around this problem, replace the %f in the formatting string of the printf function with %e, which will deplay the number to great precision with scientific notation. I did not have it as %e because I find scientific notation extremely annoying. 7 i% e( J1 G/ p. L$ ]6 D: Y5 h) I( H& x6 C
C0B40000: E4 A8 ^4 K: A2 V' z9 n
0xC0B40000, -5.6250000 G" `6 F/ U9 |! g8 a3 ]. q
43A4B200 ) C; J6 K$ [- d Y6 Y: B h0x43A4B200, 329.390625 c X9 H% T4 O
00000000" ?3 Z' h# m" B) M9 s. D
0x00000000, 0.000000 " s2 m. J. K- S+ ]- R* c$ K; t: S80000000 - @# v- N- R! H, t3 ?& F0x80000000, -0.000000 , s0 `- g. c1 y7f800000$ v2 n6 F0 R4 v
0x7F800000, inf - A" N( ~, W7 s( x* Eff800000 9 n, v; H1 G9 k1 M% ]: E: w0xFF800000, -inf - F v9 h, X4 j7fffffff 8 e8 d8 [% I4 r/ o0x7FFFFFFF, NaN1 L/ Z" {6 z3 N, g
ffffffff5 [! p# K. K) B" _. v
0xFFFFFFFF, NaN # i/ i5 B; U, G6 z7f81A0239 } }5 g# t; z/ r5 Q5 V3 |
0x7F81A023, NaN! m" }# w7 k" {9 S
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Float 2 Hex 1 a g4 m8 M, K! F- e- P#include <stdio.h>: {8 k0 C' N; c2 U
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int main() m7 P/ s# H0 L9 p2 H1 m% R
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float theFloat;6 U g- B( @4 k! g/ j( o
while (1) { }( \2 l1 g0 \ \& v3 m4 J8 e
scanf("%f", &theFloat);) K* C- T% c; ?' ~9 p2 v& ~
printf("0x%08X, %f\n", *(int *)&theFloat, theFloat); $ z$ y- R U$ M }# x- Y3 u. l; ]
return 0;0 v) M4 z6 R) v$ Y4 t
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This is a slight modification of the "Hex 2 Float" program. The exception is it reads in a floating point number. Just like and outputs the hexadecimal form plus the floating point number. Again I include a sample run of this program, confirming the results of the example problems I covered earlier in this text, along with some other simple cases. Notice the hexadecimal representation of 0.2. . o+ d( h! ~' F! D& b2 A- R2 I- w3 e0 w# D! d# Q$ R! `- r6 h, L) h
-5.625 ( M8 z9 H% {& w \2 |0xC0B40000, -5.6250004 X. X! s$ q N7 \ r
329.390625 1 i2 l* p, U4 `% S( c- A+ g. m" X0x43A4B200, 329.390625# S1 d# E& t1 b9 L
0, ^0 q% R7 I# U( d$ J
0x00000000, 0.000000 - V( m. \- Y+ E" C0 _-0; s0 T/ c$ v# y2 e1 Y! p+ m' b& s
0x80000000, -0.000000 - S; u) o4 g! W8 ~.2 & `3 P+ W) \ G5 ?0x3E4CCCCD, 0.200000 t( [/ i7 P& D' k( b.5: Y5 H# S% }9 l% K& `1 B1 `4 G
0x3F000000, 0.500000/ k7 j) \9 ], a7 {) m, f2 _/ K! J
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0x3F800000, 1.000000作者: xll 时间: 2010-2-16 01:36
多谢楼主，我来看看 [s:45]作者: xll 时间: 2010-2-16 01:37
正需要 [s:42]