我认为这是tanennmbaum在书中的数据结构中的程序
该代码是错误的,因为outrat->numerator/=a;
是无效的
命令,因为它没有分配任何值,如果我需要更正我
是错的。请更正代码。
#include <iostream.h>
#include <conio.h>
#include <stdio.h>
struct rational
{
int numerator;
int denominator;
};
void reduce(struct rational *inrat, struct rational *outrat)
{
int a, b, rem;
if (inrat->numerator > inrat->denominator)
{
a = inrat->numerator;
b = inrat->denominator;
}
else
{
a = inrat->denominator;
b = inrat->numerator;
}
while (b != 0)
{
rem = a % b;
a = b;
b = rem;
}
outrat->numerator /= a;
outrat->denominator /= a;
}
int equal(struct rational *rat1, struct rational *rat2)
{
struct rational r1, r2;
reduce(rat1, &r1);
reduce(rat2, &r2);
if (r1.numerator == r2.numerator && r1.denominator == r2.denominator)
return 1;
return 0;
}
int main()
{
struct rational rn1, rn2;
int k;
rn1.numerator = 5;
rn1.denominator = 10;
rn2.numerator = 1;
rn2.denominator = 2;
k = equal(&rn1, &rn2);
cout << k;
getch();
return 0;
}
最佳答案
将相关行替换为:
outrat->numerator = inrat->numerator / a;
outrat->denominator = inrat->denominator / a;