hi

~Rick <[EMAIL PROTECTED]> wrote:          At Thursday 11/22/2007 09:46 AM, you 
wrote:
>Hi,everyone .When I try to change decimal number into other type numbers.
> I write it like as follow:
>
>
> int main()
>{
> int integer;
> int a[100] = {0};
> static double account = 0;
> int i ;
> double ii = 0;
> double number = 0;
> double digit;
> printf("Enter one decimal number");
> scanf("%d",&integer);
> printf("Enter one number as digit you want to change to");
> scanf("%lf",&digit);
>
> while (number <= integer)
> {
> number = pow(digit, account);
> account += 1;
> }
>
>
>
> for(i = account;i >= 0;i--)
> {
> ii = i;
> a[i] = integer / pow(digit, ii);
> integer = integer - pow(digit, ii) * a[i];
> }
> printf("\n%d",account); //I am puzzled here
>
> printf("\nThe %.0f number you need is ", digit);
> for(i = (account - 2);i >= 0;i--)
> {
> if (a[i] == 10)
> printf("A");
> else if (a[i] == 11)
> printf("B");
> else if (a[i] == 12)
> printf("C");
> else if (a[i] == 13)
> printf("D");
> else if (a[i] == 14)
> printf("E");
> else if (a[i] == 15)
> printf("F");
>
> if ((a[i] == 10)||(a[i] == 11)||(a[i] == 12)
> ||(a[i] == 13)||(a[i] == 14)||(a[i] == 15))
> continue;
>
> printf("%d", a[i]);
> }
>
> return 0;
>}
>
>
> But I am puzzled by the VALUE 
> of account .I print it after the first while 
> loop , but its value still keeps as 0.I think it should be some number != 0
> And if I want to print the result I have to 
> initialize i as (account - 2) ,otherwise
> it will print two 0 before the right answer .Why ?
>

You have defined "account" to be of type double. Try

printf("\n%52.f",account); //I am puzzled here

~Rick

>
>---------------------------------
> @yahoo.cn ÐÂÓòÃû¡¢ÎÞÏÞÁ¿£¬¿ìÀ´ÇÀ×¢£¡
>
>[Non-text portions of this message have been removed]
>
>
>
>To unsubscribe, send a blank message to 
><mailto:[EMAIL PROTECTED]>.
>Yahoo! Groups Links
>
>
>



                         

       
---------------------------------
Be a better sports nut! Let your teams follow you with Yahoo Mobile. Try it now.

[Non-text portions of this message have been removed]

Reply via email to