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int divide(a,b) {
if (b == 0) print "Invalid";
else if (a < b) return 0;
else return(1 + (divide(a-b, b));
}

Test cases:
1. a = 0
2. b = 0
3. a < b
4. b =1
5. b > a

ask, Is a is upper value?
incase he says yes then
ans:
a - b != a

Note: a value cannot be devide by 0. i am figuring out b is 0 or not

Hi

How about these ? Please let me know if there are more.

1. is a>b ?
2. is b!=0 ?
3. are a and b of same or different types ?
4. If they are of different types, do they
atleast extend the Number class ?

if (b == 0) {
cout << "cannot divide by 0!" << endl;
return 0;
}
rest = a;
result = 0;
while (rest*a > 0) {
rest = a - b;
result = result + 1;
}
return result

private static void divideTwoNumbersWithoutDividingOperator(int x, int y)
{
int res = 0;
int rem = 0;
int sign = 1;

//check the sign
sign = (x < 0) ? -1 : 1;

x = x * sign;//change the sign to a positive number

if (x < y)
{
rem = x;
}

while ((x - y) > 0)
{
rem = x - y;
x = x - y;
++res;
}

res = res * sign;//reset the sign

Console.WriteLine("x/y = " + res);
Console.WriteLine("x%y = " + rem);
}

Try each combination of:
a=0,a=1,a=-1,a=null(if possible) and b=0,b=1,b=-1,b=null
The above cases covers equivalence classes of -ve result,+ve result and invalid result(division by zero). These even follow boundary value analysis.

Do performance testing by setting a=2^32 and b=2

Floating point result converted to int result: Try a=3,b=2 o/p=1 and a=2,b=3 o/p=0
a>b
a<b
and a=b(all covered in one of the above test cases)

#include<stdio.h>
int div(int a,int b);
main()
{
int q,w,x;
scanf("%d %d",&q,&w);
x=div(q,w);
printf("%d",x);
}
int div(int x,int y)
{
int z=x,count=0;
while(z>=y)
{
z-=y;
count++;
}
return count;
}

Also take care if b is negative.
if(b<0) call res=divide(a,-b) & if a is negative call res=divide(-a,b). The answer is -res.
If both are negative call divide(-a,-b)

int divide(a,b) {
          if (b == 0) print "Invalid";
           else if (a < b) return 0;
           else return(1 + (divide(a-b, b));
}