Walter Blyss的部落格

Perl 的資料型態:
Perl 的資料型態，不像C/C++, Java等....的程式語言，有int, float, double, char...，
它的資料型態，有

有了Perl, 但如何能知道本身Perl的版本呢?
很簡單，perl -v 就搞定了，如下:
root@walter:/opt/OPT/testperl# perl -v
This is perl, v5.10.1 (*) built for i486-linux-gnu-thread-multi
Copyright 1987-2009, Larry Wall
Perl may be copied only under the terms of either the Artistic License or the
GNU General Public License, which may be found in the Perl 5 source kit.
Complete documentation for Perl, including FAQ lists, should be found on
this system using "man perl" or "perldoc perl".  If you have access to the
Internet, point your browser at http://www.perl.org/, the Perl Home Page.

Perl, 到底Perl是什麼?
一定很多人有如此疑問，就連當初剛學習的我也不例外，
其實，Perl和其他程式語言一樣，也就是它，也是一種程式語言(這句好像是廢話XD)，
當初開發Perl的Larry Wall，是為了方便處理報表工作而開發的，

純參考喔， 加油^^
#include <stdio.h>
#include <stdlib.h>
int Input(int *arr, int loop)
{
        int i;
        for (i = 0; i < loop; i++) {
                printf("%2d input number:", i + 1);
                scanf("%d", &arr[i]);

純參考喔^^加油
[root@walter yahoo-code]# cat test0616.c
#include <stdio.h>
#include <stdlib.h>
/*****************************************************************************/
#define ShareFun \
        for (i = 0;i < 3; i++) { \
            arr3[i] = Fun(arr1[i], arr2[i]); \
        }
/*****************************************************************************/
int (*Fun)(int num1, int num2);
int AddFun(int num1, int num2)
{ int num3 = num1 + num2; return num3; }
int SubFun(int num1, int num2)
{ int num3 = num1 - num2; return num3; }
int TimeFun(int num1, int num2)
{ int num3 = num1 * num2; return num3; }
int PriArr(int *arr, int length)
{    
    int i;
    for (i = 0;i < length; i++) {
        printf("arr3[%d] = %2d ", i, arr[i]);
    }
    printf("\n=======================================\n");
}
int main ()
{
    int i;
    int arr1[3] = {1, 2, 3};
    int arr2[3] = {3, 2, 1};
    int arr3[3] = {0, 0, 0};
    /* Call Add Fun    */
    printf("Call Add Fun:\n");
    Fun = AddFun;
    ShareFun
    PriArr(arr3, sizeof(arr3)/sizeof(arr3[0]));
    
    /* Call Sub Fun    */
    printf("Call Sub Fun:\n");
    Fun = SubFun;
    ShareFun
    PriArr(arr3, sizeof(arr3)/sizeof(arr3[0]));
    
    /* Call Time Fun */
    printf("Call Time Fun:\n");
    Fun = TimeFun;
    ShareFun
    PriArr(arr3, sizeof(arr3)/sizeof(arr3[0]));
    
    return 0;
}

#!/bin/bash
PATH=/bin:/sbin:/usr/bin:/usr/sbin:~/bin
export PATH
add="add"
sub="sub"
mul="mul"
div="div"
while true
do
        j=0
        answer=0
        echo "=========="
        echo "請以add 取代+"
        echo "請以sub 取代-"
        echo "請以mul 取代*"
        echo "請以div 取代/"
        echo "=========="
        read -p "Please Input number(ex.add 1 2 3):" A[0]
        for i in ${A[0]}
        do
                if [ $j == 0 ]; then
                        option=$i
                        j=$(($j+1))
                else
                        tmp[$j]=$i
                        j=$(($j+1))
                fi
        done
        case $option in
                $add)
                        answer=0
                       for ((k=1; k<$j; k++))
                        do
                                answer=$(($answer+$((tmp[$k]))))
                        done
                        ;;
                $sub)
                        answer=${tmp[1]}
                        for ((k=2; k<$j; k++))
                        do
                                answer=$(($answer-$((tmp[$k]))))
                        done
                        ;;
                $mul)
                        answer=${tmp[1]}
                        for ((k=2; k<$j; k++))
                        do
                                answer=$(($answer*$((tmp[$k]))))
                        done
                        ;;
                $div)
                        answer=${tmp[1]}
                        for ((k=2; k<$j; k++))
                        do
                                answer=$((($answer)/$((tmp[$k]))))
                        done
                        ;;
                *)
                        echo "Please input add, sub, mul or div, thanks."
                        continue
                        ;;
        esac
        echo "The answer is $answer"
done
exit 0
執行結果如下:

a.txt 如下:
ABC
87888
584587
232323
ABC
78787
12121
898989
ABC
45454
454547
898989
程式碼如下: