have to identify odd num and print them

[ravali]

#include<stdio.h>
#define TRUE 1

int formOddInt(int n);

int main() {

	int num;

	do {

			printf("Enter a positive integer\n");
			scanf("%d", &num);
		} while (num <= 0);

	printf("Num is = %d \n The new number is %d \n", num,formOddInt(num));

	return TRUE;
}

 int formOddInt(int n)
{
int i=0,d[5],count=0;
while(n>0) {
d[i]=n%10;
n=n/10;
i++ ;
count++ ;
}
for(i=0;i<count;i++)
printf("count=%d\td[i]=%d\n",count,d[i]);
for(i=count;i>0;i--)
{
 if((d[i]%2)!=0)
 {
	return(d[i]);
   }
}
   }

[johny10151981]

You did not tell your problem. You should not expect that people would read your code and figure out your problem.

[vijay6]

Hey ravali, the logical error is in line number 32 in your code. Inside the for loop assign value for 'i' as 'count - 1' instead of 'count'.

for( i = count - 1 ; i > 0 ; i-- ) // Line number 32

And after 38th line write 'return -1' which means if input number don't have odd number then it'll print -1.

return -1; // After line number 38

Note: Next time post/ask your question along with your code.

[gaurav moury]

hi m gaurav..

// identify odd no. and print them
#include<iostream.h>
void main()
{ int a[10],b;
cout<<"enter how many nos.";
cin>>b;
for(int i=0;i<=b;i++)
{ cout<<"enter "<<i+1<<"element";
  cin>>a[i];
}
cout<<"the odd nos. are  ";
for(i=0;i<=b;i++)
{if(a[i]!%2)
cout<<a[i];
}
}

//i wish this code would b understood by u..

[fishirboy]

Ya C++ with #include <iostream> but some people learn different and some times there way may be obsolete but they know it best and can use it the same as the new stuff, it's just harder.

[johny10151981]

iostream.h is obsolete, but in South Asia region people start with turbo c++ 3.0, a 20 years old compiler. I believe their teacher should not suggest their pupil this compiler......