loop and switch.. help:)

[houserocks13]

im quite confused with loops and switch.. will you please help me:)
i have 3 problems..:)

1.Write a program that will ask for a long number and count how many digits are there in the number.
ex. Enter num: 10854
Output: 5

2.Write a program that will ask for a long number and count how many digits in the number are even and how many are odd.
ex. Enter num: 80572
Output: 3 digits are even
2 digits are odd

3.Write a program that asks for an input integer and displays the digits in reverse.
ex. Enter number: 3562 ex. Enter number: 10240
Output: 2653 Output: 04201

[horace1]

implement and test part 1 first then deal with the others when that is working, i.e.

1.Write a program that will ask for a long number and count how many digits are there in the number.
ex. Enter num: 10854
Output: 5

probably the simplest way is to read the number into a char array and use the string processing functions in <string.h>

[houserocks13]

hehehe.. we havent tackeld arrays yet.. only loops.. so we are supposed to loop for this program.. my syntaz doesnt work..=( it doesnt display the 0 wahhh...

[r035198x]

hehehe.. we havent tackeld arrays yet.. only loops.. so we are supposed to loop for this program.. my syntaz doesnt work..=( it doesnt display the 0 wahhh...

Well have you done string manipulation then? You'd need to read in the number as a string and access each character of the string for manipulation using string methods

[houserocks13]

not yet=(

for this number
2.Write a program that will ask for a long number and sum the even and odd digits in the number.

my prof changed it

i have this code:
#include <iostream.h>


void main()
{
int num, r, evensum=0, oddsum=0;

cout<<"Enter number: "<< endl;
cin>>num;

r=num/10;
while (num>0)
{
if (r%2==0)
evensum=evensum +r;
else
oddsum=oddsum+r ;
num--;
}
cout<<"even is: "<<evensum< < endl;
cout<< "odd is "<< oddsum<<endl;
}


but it wont print the sum.. huhu=(

[horace1]

you had a few errors

#include <iostream.h>

int main()
{
int num, r, evensum=0, oddsum=0;

cout<<"Enter number: "<< endl;
cin>>num;

while (num>0)           // loop thru all digits
{
if (num%2==0)           // check if even number
evensum=evensum+1;
else
oddsum=oddsum+1;
num=num/10;             // check next digit
}
cout<<"even is: "<<evensum<< endl;
cout<< "odd is "<< oddsum<<endl;
}

you should now be able to extend this to solve your other problems

[koder]

implement and test part 1 first then deal with the others when that is working, i.e.

1.Write a program that will ask for a long number and count how many digits are there in the number.
ex. Enter num: 10854
Output: 5

probably the simplest way is to read the number into a char array and use the string processing functions in <string.h>

better use '%' (Modulus operator) ok?

[horace1]

better use '%' (Modulus operator) ok?

Usually there are serveral ways of implementing an algorithm which vary in simplicity, efficency (run time or memeory required), etc.
For example, using a char array to count the number of even and odd digits entered at the keyboard

#include <iostream>
using namespace std;

int main()
{
int evensum=0, oddsum=0;
char number[50];
cout<<"Enter number: "<< endl;
cin>>number;                           // read in as characters

for(int i=0; number[i] != '\0'; i++)   // loop thru all digits
  if ((number[i] & 1) == 0)            // check if even number
       evensum=evensum+1;
   else
       oddsum=oddsum+1;
cout<<"even is: "<<evensum<< endl;
cout<< "odd is "<< oddsum<<endl;
}

This uses the & operator to determine if a character is odd or even.
It can deal with a larger number of digits than reading an int and at the end of this code we still have the original data in number[].
It all depends what the specification is.