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A simple program help???

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Zangetsu

I am the master of my Fate.
Hi Guys, :oops:

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?:cry:
 

Vishal Patil

Linux all the way
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
 
Last edited:

lucifer_is_back

Journeyman
Code:
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
 
OP
Zangetsu

Zangetsu

I am the master of my Fate.
Ohhh! so simple, Thanks Vishal :oops:

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

Vishal Patil

Linux all the way
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++;
}
 
OP
Zangetsu

Zangetsu

I am the master of my Fate.
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...:-D
 
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