5.3 Solutions

```
mystery
```
```
a   b
```
```
int
```
```
int mystery(float a, float b)
```

int mystery (float a, float b)
{
   int value;
   if (a < b)
      value = -1;
   else if (a > b)
      value = 1;
   else
      value = 0;
   return value;
}

int mystery (float a, float b)
{
   if (a < b)
      return -1;
   else if (a > b)
      return 1;
   else
      return 0;
}

char first(char a, char b)
{
   if(a < b)
      return a;
   else
      return b;
}

Returns the smaller parameter.

float second(float a, float b)
{
   float answer;
   if(a < b)
      answer = a - b;
   else
      answer = b - a;
   return answer;
}

Returns the negative difference of the parameters.

c) is invalid since println is a void method. It doesn't return anything to store in x.
d) is valid, but pointless since the value of f(x) is lost.

float largest(float a, float b, float c)
{
   if (a > b && a > c)
      return a;
   else if (b > c)
      return b;
   else
      return c;
}

int gcd (int x, int y)
{
   int divisor = 1, small = abs(x);
   if (abs(y) < abs(x))
      small = abs(y);
   for (int i = 2; i <= small; i++)
      if (x % i == 0 && y % i == 0)
         divisor = i;
   return divisor;
}