函数大全(k开头)

韩冰

< align=center><FONT color=#0000ff size=3><B><FONT color=#cc0000>函数大全(k开头)</FONT></B></FONT>
1 G. y0 F$ U# p5 O</P>
<><FONT color=#ff0000>函数名: kbhit </FONT>
功 能: 检查当前按下的键
用 法: int kbhit(void);
程序例: </P>
: w" n! {) v- Q: k2 g9 U' L<><FONT color=#0000ff>#include <CONIO.H></FONT></P>
' o) L  O6 `* _# G/ U" l, ?<><FONT color=#0000ff>int main(void)
{
cprintf("ress any key to continue:");
while (!kbhit()) /* do nothing */ ;
cprintf("\r\nA key was pressed...\r\n");
return 0;
' t% E( Y# d8 P} </FONT>
</P>
<><FONT color=#ff0000>函数名: keep </FONT>
( e0 h8 g1 Q% u7 S功 能: 退出并继续驻留
# V3 b4 i9 c: I0 w0 a6 B$ b, n用 法: void keep(int status, int size);
程序例: </P>
<><FONT color=#0000ff>/***NOTE:
4 H, z- l% T& C# vThis is an interrupt service routine. You
can NOT compile this program with Test
Stack Overflow turned on and get an
9 U/ E5 x; c4 M" _+ k" P3 X" Kexecutable file which will operate
0 t5 @6 P5 v# z$ j: Dcorrectly. Due to the nature of this
function the formula used to compute
the number of paragraphs may not
! Q6 n) c$ J( o2 X4 @necessarily work in all cases. Use with
care! Terminate Stay Resident (TSR)
, S6 d/ ^$ M0 A* D+ R# V4 pprograms are complex and no other support
for them is provided. Refer to the
, W! S0 j! s) f1 v. LMS-DOS technical documentation
2 r/ \: b/ a1 Zfor more information. */
* D2 l+ U; O5 X9 ~  ^#include <DOS.H>
' R# U( l. ^7 }$ l/ Q; Z+ L/* The clock tick interrupt */
#define INTR 0x1C
/* Screen attribute (blue on grey) */
#define ATTR 0x7900 </FONT></P>
<><FONT color=#0000ff>/* reduce heaplength and stacklength
to make a smaller program in memory */
! I$ t% G' x& [' ~2 U5 j1 |1 rextern unsigned _heaplen = 1024;
extern unsigned _stklen = 512; </FONT></P>
<><FONT color=#0000ff>void interrupt ( *oldhandler)(void); </FONT></P>
. t$ u' R0 m2 D2 t. `% z5 i' r) H<><FONT color=#0000ff>void interrupt handler(void)
  \/ W3 t3 J5 O: C{
2 D+ x- T% c& a( z7 |% Dunsigned int (far *screen)[80];
static int count; </FONT></P>
<><FONT color=#0000ff>/* For a color screen the video memory
" R' i. U) e/ D4 z- L/ Eis at B800:0000. For a monochrome
system use B000:000 */
screen = MK_FP(0xB800,0); </FONT></P>
3 A8 H! d: l+ a/ z8 O: }5 _7 `<><FONT color=#0000ff>/* increase the counter and keep it
within 0 to 9 */
count++;
; e: G! V5 u. {4 dcount %= 10; </FONT></P>
<><FONT color=#0000ff>/* put the number on the screen */
screen[0][79] = count + '0' + ATTR; </FONT></P>
<><FONT color=#0000ff>/* call the old interrupt handler */
oldhandler();
5 m0 S8 @4 E  f  B4 U, n. E9 i4 [6 H/ x} </FONT></P>
<><FONT color=#0000ff>int main(void)
  D% w3 F7 j5 h$ R7 A! a  {{ </FONT></P>
<><FONT color=#0000ff>/* get the address of the current clock
: F) Q3 t9 P. ktick interrupt */
; @" c# Q! k$ |) e3 Aoldhandler = getvect(INTR); </FONT></P>
<><FONT color=#0000ff>/* install the new interrupt handler */
setvect(INTR, handler); </FONT></P>
. l& `: d+ q; k<><FONT color=#0000ff>/* _psp is the starting address of the
program in memory. The top of the stack
( A5 v  a: j+ p! qis the end of the program. Using _SS and
_SP together we can get the end of the
0 R! T, v3 E5 g) B2 s9 Y* j/ Ustack. You may want to allow a bit of
saftey space to insure that enough room
is being allocated ie:
(_SS + ((_SP + safety space)/16) - _psp)
# E! J) f) G* C3 L*/
keep(0, (_SS + (_SP/16) - _psp));
& m1 j0 N7 t7 O0 Vreturn 0;
}
! b6 y  w' Q3 b</FONT></P>


1 c* s( G/ [+ i: o) P+ A7 i<><FONT color=#ff0000>函数名: kbhit </FONT>
功 能: 检查当前按下的键
  n! |( E; E- X2 _7 \用 法: int kbhit(void);
; }' p* D5 }  l  q8 K% e程序例: </P>
<><FONT color=#0000ff>#include <CONIO.H></FONT></P>
' E7 X8 d" ?+ F* }$ H<><FONT color=#0000ff>int main(void)
$ _* N* ?7 X6 s) y5 ~: v- Q{
cprintf("ress any key to continue:");
. n) ]2 J& U+ R4 ewhile (!kbhit()) /* do nothing */ ;
cprintf("\r\nA key was pressed...\r\n");
return 0;
}
- o1 ]+ H, O; z

# Q* \: q" M9 V5 ?  k1 v. I</FONT></P>
<><FONT color=#ff0000>函数名: keep </FONT><FONT color=#0000ff>
<FONT color=#000000>功 能: 退出并继续驻留
: f1 Q, _( k  V) e- e用 法: void keep(int status, int size);
- M+ I# v2 n% I' u; ?3 m1 ]程序例: </FONT></FONT></P>
<><FONT color=#0000ff>/***NOTE:
This is an interrupt service routine. You
6 a7 t7 L2 ~6 _' V$ w9 u: hcan NOT compile this program with Test
Stack Overflow turned on and get an
! s; J, [4 p4 d8 K+ l* uexecutable file which will operate
correctly. Due to the nature of this
  z; ~$ S! U& W4 Y# Y2 b) Cfunction the formula used to compute
: c5 j, ~  u( c! Nthe number of paragraphs may not
/ {/ z/ W+ q/ A. Q4 K: D" O1 ~necessarily work in all cases. Use with
care! Terminate Stay Resident (TSR)
programs are complex and no other support
for them is provided. Refer to the
$ C/ y( q+ g6 T+ q2 I& cMS-DOS technical documentation
for more information. */
$ e/ p. H8 z2 i3 W% g#include <DOS.H>
4 j5 B5 V7 b; s0 o$ R/* The clock tick interrupt */
#define INTR 0x1C
) U8 a) ]. P  o# G/* Screen attribute (blue on grey) */
#define ATTR 0x7900 </FONT></P>
* @- E3 U/ Q( P# x4 v+ o<><FONT color=#0000ff>/* reduce heaplength and stacklength
to make a smaller program in memory */
extern unsigned _heaplen = 1024;
3 t6 g9 X/ l& H% Oextern unsigned _stklen = 512; </FONT></P>
! I( ^  H! p! O) @<><FONT color=#0000ff>void interrupt ( *oldhandler)(void); </FONT></P>
- z7 q% K, t) J. O7 p<><FONT color=#0000ff>void interrupt handler(void)
/ O0 e, _2 X" K% A" `{
5 O' r, o6 R& k' l) L2 V% Punsigned int (far *screen)[80];
# G& y/ I2 L% W* h2 Ostatic int count; </FONT></P>
0 m, W; ]! c% S0 P" U. u<><FONT color=#0000ff>/* For a color screen the video memory
is at B800:0000. For a monochrome
. D( W/ j: |, n! |7 e2 Gsystem use B000:000 */
4 P  q' E7 i, e1 V- L1 K* o# Rscreen = MK_FP(0xB800,0); </FONT></P>
<><FONT color=#0000ff>/* increase the counter and keep it
# R2 k& u) {/ ]8 ~, r3 O) qwithin 0 to 9 */
count++;
count %= 10; </FONT></P>
) ?2 I- t0 g9 S$ T2 B) K<><FONT color=#0000ff>/* put the number on the screen */
screen[0][79] = count + '0' + ATTR; </FONT></P>
<P><FONT color=#0000ff>/* call the old interrupt handler */
oldhandler();
6 u# e6 [9 E$ p} </FONT></P>
<P><FONT color=#0000ff>int main(void)
{ </FONT></P>
( K8 j) C  D2 U: G: U/ E<P><FONT color=#0000ff>/* get the address of the current clock
tick interrupt */
( @1 A# q* @7 h. r2 @7 Foldhandler = getvect(INTR); </FONT></P>
( F7 g% y- ~! ~5 y! h<P><FONT color=#0000ff>/* install the new interrupt handler */
setvect(INTR, handler); </FONT></P>
<P><FONT color=#0000ff>/* _psp is the starting address of the
7 R6 F# g$ B) T, U5 Aprogram in memory. The top of the stack
+ G* m5 |0 @( L) H9 K& fis the end of the program. Using _SS and
! a, f( T; v7 I9 s9 Q+ E_SP together we can get the end of the
  h: }- b! \9 |6 ostack. You may want to allow a bit of
saftey space to insure that enough room
is being allocated ie:
(_SS + ((_SP + safety space)/16) - _psp)
*/
  F0 l! a0 E7 W4 t0 I8 M( L# Kkeep(0, (_SS + (_SP/16) - _psp));
4 I; U7 S' K0 `! k/ yreturn 0;
: u$ a5 E7 A" N" O  W}</FONT></P>