A simple program help???

[Zangetsu]

Hi Guys,

If have a problem in solving this & i googled but cudn't find ne solution...

if integer a = 12345. write a program (without using any in-built function) in C++/C# or
VB.NET. to find the length of integer a (obviously in this case it is 5).remember if its array then its simple.but its an integer datatype.

pls solve my problem?

[Vishal Patil]

int a=12345;
int count = 0;
while(a !=0 )
{
a = a/10; //removes the last digit of the number; or better a /= 10
count++;
}


after the loop is executed a will be zero count will contain the number of digits in the number. if you want back a just use int b = a; //before the loop a =b; //and after the loop

[lucifer_is_back]

int a=12345;
if(a<0)a=-a;
int count = 0;
while(a !=0 )
{
a = a/10; //removes the last digit of the number; or better a /= 10
count++;
}

check for negative

[Vishal Patil]

does work. the condition is a ! = 0; not a < 0

[Zangetsu]

Ohhh! so simple, Thanks Vishal

ok...if i want 2 reverse the digit then like o/p= 54321 in the same program

[Vishal Patil]

int a=12345;
if(a<0)a=-a;
int count = 0;
int remainder, reverse=0; // ----change over here
while(a !=0 )
{

remainder = a%10; // to read the last digit -----change

a = a/10; //removes the last digit of the number; or better a /= 10

reverse = (reverse * 10) + remainder; //this statement should come before count++ else the value will be 543210

count++;
}

[Zangetsu]

Originally Posted by Vishal Patil

int a=12345;
if(a<0)a=-a;
int count = 0;
int remainder, reverse=0; // ----change over here
while(a !=0 )
{

remainder = a%10; // to read the last digit -----change

a = a/10; //removes the last digit of the number; or better a /= 10

reverse = (reverse * 10) + remainder; //this statement should come before count++ else the value will be 543210

count++;
}

Oh..thank u very much...but i think for the above program count is not needed...