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If we know logic how do we write code for it
For an example
quotient = 987 / 10 --> 98
remainder = 987 % 10 --> 7 -> first number
quotient = 98 / 10 --> 9 --> third number
remainder = 98 % 10 --> 8 --> second number
first ->second -> third = 7 8 9
I do not want the code, I want to find out way how I can write my code if I know the logic.
If you know your logic, then the only way you can code it is by starting to type, I guess?
Normally I know How do the Reverse Number. The problem is that how do I implement the program ?Quote:
Originally Posted by Zeus_
Code:#include<stdio.h>
int main ()
{
int number = 987;
int reminder;
int reverse = 0;
reminder = number % 10; // remainder = 987 % 10 --> 7 -> first number
number = number / 10; // quotient = 987 / 10 --> 98
reminder = number % 10; //remainder = 98 % 10 --> 8 --> second number
number = number / 10; // quotient = 987 / 10 --> 98 quotient = 98 / 10 --> 9 --> third number
return 0;
}
Do you know about while loops?
Now would be a good time to find out.
As in, while you have some number > 0, you do your % and / thing.
I am using % and / operatorQuote:
Originally Posted by Salem
What will be the algorithm inside of the while loopCode:#include<stdio.h>
int main ()
{
int number = 987;
int reminder;
int reverse = 0;
while ( number > 0 )
{
reminder = number % 10; // remainder = 987 % 10 --> 7 -> first number
number = number / 10; // quotient = 987 / 10 --> 98
reminder = number % 10; //remainder = 98 % 10 --> 8 --> second number
number = number / 10; // quotient = 987 / 10 --> 98 quotient = 98 / 10 --> 9 --> third number
}
return 0;
}
Try it with only one % and one / inside the loop.
Maybe just this to begin with.
Code:while ( number > 0 ) {
printf("Number=%d\n", number);
number = number / 10;
}
Quote:
Originally Posted by Salem
Number=987Code:#include<stdio.h>
int main ()
{
int number = 987;
int reminder;
int reverse = 0;
while ( number > 0 ) {
reminder = number % 10; // remainder = 987 % 10 --> 7 -> first number
printf("Number=%d\n", number);
number = number / 10;
// reminder = number % 10; // remainder = 987 % 10 --> 7 -> first number
printf("reminder =%d\n", reminder);
}
return 0;
}
reminder =7
Number=98
reminder =8
Number=9
reminder =9
> reminder =7
> reminder =8
> reminder =9
Good, now can you put them all into reverse ?
I tried but I do not understand how to do thatQuote:
Originally Posted by Salem
What about
reverse = reverse * 10 + reminder
I got itQuote:
Originally Posted by Salem
Reverse number is = 789Code:#include<stdio.h>
int main ()
{
int number = 987;
int reminder;
int reverse = 0;
while ( number > 0 ) {
reverse = reverse * 10 + number % 10;
number = number / 10;
}
printf("Reverse number is = %d \n", reverse);
return 0;
}
One little thing, thou... this:Quote:
Originally Posted by Parth12
Will result in 0!Code:#include <stdio.h>
int main( void )
{
int number = -987;
int reverse = 0;
while ( number > 0 )
{
reverse = reverse * 10 + number % 10;
number = number / 10;
}
printf( "Reverse number is = %d \n", reverse );
}