Objective

Reduce the fraction held in r in place by dividing num and den by their greatest common divisor, with the sign carried by num so den ends up strictly positive.

Steps

• r points at one rational, declared as typedef struct rational { int num; int den; } rational;. Write the reduced fraction back into that same struct, the function returns nothing.
• Divide num and den by the greatest common divisor of their magnitudes. 6/8 reduces to 3/4, and a fraction already in lowest terms comes back unchanged.
• The sign belongs to the numerator. After reduction den is strictly positive, so 3/-4 becomes -3/4 and two negatives cancel: -6/-8 becomes 3/4.
• A num of 0 reduces to 0/1 whatever den held. Zero has no denominator worth keeping, so Grademe pins it to 1.
• A fraction that divides evenly keeps a denominator: 8/4 becomes 2/1, never 2 on its own.
• No library function is available. Write the gcd yourself, Euclid's remainder loop is short enough.
• Both fields fit in an int, and so does the reduced result. Inputs stay inside the range where -num and -den are representable, so INT_MIN never reaches you.
• den is never 0. A zero denominator sits outside the domain of the function, Grademe never calls you with one, and it owes you no special case.

Objective

Reduce the fraction held in r in place by dividing num and den by their greatest common divisor, with the sign carried by num so den ends up strictly positive.

Steps

• r points at one rational, declared as typedef struct rational { int num; int den; } rational;. Write the reduced fraction back into that same struct, the function returns nothing.
• Divide num and den by the greatest common divisor of their magnitudes. 6/8 reduces to 3/4, and a fraction already in lowest terms comes back unchanged.
• The sign belongs to the numerator. After reduction den is strictly positive, so 3/-4 becomes -3/4 and two negatives cancel: -6/-8 becomes 3/4.
• A num of 0 reduces to 0/1 whatever den held. Zero has no denominator worth keeping, so Grademe pins it to 1.
• A fraction that divides evenly keeps a denominator: 8/4 becomes 2/1, never 2 on its own.
• No library function is available. Write the gcd yourself, Euclid's remainder loop is short enough.
• Both fields fit in an int, and so does the reduced result. Inputs stay inside the range where -num and -den are representable, so INT_MIN never reaches you.
• den is never 0. A zero denominator sits outside the domain of the function, Grademe never calls you with one, and it owes you no special case.