How to print All zeros in C? - c

Question

I want to print all the zeros in the input using this code.
for example:
input: 10000 -------------- output: 0000
but I get this:
input: 10000 -------------- output: 0
this is the source code in C99:
#include<stdio.h>
int main() {
int co = 0, inputNumber, i, j, prod = 1, number = 0, ten = 10, newValue;
int numberOfInputs;
scanf("%d", &numberOfInputs);
for(i = 1; i <= numberOfInputs; i++) {
scanf("%d", &inputNumber);
for(j = 1; j < inputNumber; j++) {
if(inputNumber % j == 0) {
prod = prod * j;
}
}
if(prod < 10000)
printf("%d", prod);
else {
newValue = prod;
while((int)prod / 10 != 0) {
prod = (int) prod/10;
co++;
number = newValue % ten;
ten *= 10;
if(co == 4) {
printf("%d \n", number);
break;
}
}
}
}
return 0;
}
Please some help guys :)

Answer 1

So basically, you want to print out the four lowest-order digits of the product with leading zeros, which is pretty simple:
printf( "%04d\n", prod % 10000 );
Cleaning up and reformatting your code a bit basically reduces to this:
#include <stdio.h>
int main( void )
{
int inputNumber, i, j, prod=1;
int numberOfInputs;
scanf( "%d", &numberOfInputs );
for(i = 1; i <= numberOfInputs; i++ )
{
scanf("%d", &inputNumber);
for(j = 1; j < inputNumber; j++ )
{
if( inputNumber % j == 0)
{
prod = prod*j;
}
}
printf( "%04d\n", prod % 10000 );
}
return 0;
}

Answer 2

Using a completely different approach, you can do this:
#include<stdio.h>
int main(){
int zeroCounter=0;
char inputNumber[256];
printf("Please enter the numerical value:");
fgets(inputNumber,256, stdin);
int j;
for(j=0; inputNumber[j]!='\n'; j++)
{
if(inputNumber[j]=='0') zeroCounter++;
}
printf("this value had %d zeros, here they are: ",zeroCounter);
while(zeroCounter!=0)
{
printf("0");
zeroCounter--;
}
printf("\n");
return 0;
}
what this code does is it sniffs the character array you input, and counts the 0 characters. then it can print them if required. no division and bitshifting shenanigans required.

Answer 3

Well your current problems are that because of the overflows, prod ends to be 0. And as 0 < 10000 you just pass through the single print: if(prod<10000)printf("%d",prod); just bypassing all your complex printing code.
In fact it overflows (to a negative value) at least at for a value as small as 84.
But that's not all. Even if prod does not overflow, you compute a number of at most 4 digits in number but print if in %d format. If number is less than 1000, you will not print the highest order 0 with that format, but you should use %04d.
Here is a minimally fixed version:
int main(){
int co=0,inputNumber,i,j,prod=1,number=0,ten=10,newValue;
int numberOfInputs;
scanf("%d",&numberOfInputs);
for(i=1;i<=numberOfInputs;i++)
{
scanf("%d",&inputNumber);
prod = 1; /* reset prod */
for(j=1;j<inputNumber;j++) {
if(inputNumber%j==0) {
prod = prod*j;
if (prod <= 0) {
break;
}
}
}
if (prod <=0) {
printf("Overflow\n");
}
else if(prod<10000) {
printf("%04d\n",prod);
}
else
{
newValue = prod;
while((int)prod/10 != 0)
{
prod = (int)prod/10;
co++;
number = newValue%ten;
ten*=10;
if(co==4)
{
printf("%04d \n",number);
break;
}
}
}
}
return 0;
}
But in fact the else part should just be:
else
{
printf("%04d \n",prod % 10000);
}