兩個常用的副程式：skip(), pause()

// for skip(), pause()

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <math.h>
// ----------------------------------------------

//   skip(-3);
void skip(int no)
{
	// limits no in 1 .. 20
	if ((no >= 1) && (no <= 20)) {// no is OK!
		for (int i=1;i<=no;i++) {
			// printf("i= %5d, no= %5d\n", i, no);

			printf("\n");
		}
	}
	else {// no is invalid!
		// printf("\n\n\n *** error in skip(), no= %5d\n", no);
		// system("pause");

		printf("\n");
	}
}// end of skip()
// ----------------------------------------------

//   pause();
void pause(void)
{
	int ch1;

	// remove type- ahead
	while (_kbhit()) {
		_getch();
	}
	// ----------------------------------------------

	printf(" Press any key to continue, [Esc] key to exit... ");
	do {
		// wait for key pressed
	} while (!_kbhit());
	ch1= _getch();
	printf("\n");
	// ----------------------------------------------

	if (ch1==0x1b) {// [Esc] key be pressed!
		exit(1);
	}

	// remove extra key pressed
	while (_kbhit()) {
		_getch();
	}
}// end of pause()
// ----------------------------------------------

void main()
{
	skip(3);

	for (int i=1;i<=20;i++) {
		skip(i);
		printf("i= %5d, i*i= %5d, sqrt(i)= %10.6lf\n", 
			i, i*i, sqrt(i));
		pause();
	}
	pause();
}// end of main()

评论

2 楼 sjgau 2007-12-20  

以下為 時間測量的輸出

// file: psc51.cpp
// for time1(), time2()

/*

ct1=   1, no=           10, sum=           55, t1= 13917990, dt=   0.0000
ct1=   2, no=           30, sum=          465, t1= 13917990, dt=   0.0000
ct1=   3, no=           90, sum=         4095, t1= 13917990, dt=   0.0000
ct1=   4, no=          270, sum=        36585, t1= 13917990, dt=   0.0000
ct1=   5, no=          810, sum=       328455, t1= 13917990, dt=   0.0000
ct1=   6, no=         2430, sum=      2953665, t1= 13917990, dt=   0.0000
ct1=   7, no=         7290, sum=     26575695, t1= 13917990, dt=   0.0000
ct1=   8, no=        21870, sum=    239159385, t1= 13917990, dt=   0.0000
ct1=   9, no=        65610, sum=  -2142598441, t1= 13917990, dt=   0.0000
ct1=  10, no=       196830, sum=  -2103713615, t1= 13917990, dt=   0.0000
ct1=  11, no=       590490, sum=  -1754143841, t1= 13917990, dt=   0.0000
ct1=  12, no=      1771470, sum=   1390803145, t1= 13917990, dt=   0.0100
ct1=  13, no=      5314410, sum=   -372987993, t1= 13918000, dt=   0.0300
ct1=  14, no=     15943230, sum=    922132129, t1= 13918030, dt=   0.0900
ct1=  15, no=     47829690, sum=   -338575121, t1= 13918120, dt=   0.2700
ct1=  16, no=    143489070, sum=   1104302137, t1= 13918390, dt=   0.8120
ct1=  17, no=    430467210, sum=    918317431, t1= 13919202, dt=   2.4130
ct1=  18, no=   1291401630, sum=  -1616479343, t1= 13921615, dt=   7.3110

*** end of program!
Press any key to continue
  */

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/timeb.h>
#include <time.h>
// ----------------------------------------------

#include "sj01.h"
#include "sj02.h"
// ----------------------------------------------

void main()
{
	int no, sum, i, ct1= 0;
	long t1;
	double dt;
	// ----------------------------------------------

	no= 10;
	skip(3);
	while (no > 0) {
		time1(&t1);

		sum= 0;
		for (i=1;i<=no;i++) {
			sum+= i;
		}
		time2(t1, &dt);
		// ----------------------------------------------

		ct1++;
		printf("ct1= %3ld, no= %12ld, sum= %12ld, t1= %8ld, dt= %8.4lf\n", 
			ct1, no, sum, t1, dt);

		if (dt > 0) {
			// pause();
		}

		if (ct1>=16) {
			// break;
		}

		no*= 3;
	}

	skip(1);
	printf("*** end of program!\n");
}// end of main()

1 楼 sjgau 2007-12-20  

將自己寫的副程式放到外面的檔案，
分開編譯

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <math.h>

#include "sj01.h"
// ----------------------------------------------

void main()
{
	skip(1);
	for (int i=1;i<=20;i++) {
		skip(i);
		printf("i= %5d, i*i= %5d, sqrt(i)= %10.6lf\n", 
			i, i*i, sqrt(i));
		pause();
	}
}// end of main()