（eden）simple struct

simple struct

Description:

Write a C program to add two fractions and display the result fraction. Your program will prompt the user to input fraction 1 and fraction 2. The numerator and denominator of each fraction are input separately by space.  See the example output below:

Enter fraction 1(numerator denominator): 1 2  
Enter fraction 2(numerator denominator): 2 5  
Result: 9/10   

 

Hint:

Hint is not available for this exercise

 

Code:

//main.c
1.#include <stdio.h>
2.#include "fract.h"
3. 
4. 
5. 
6.int main() {
7.    int num1, deno1, num2, deno2;
8.    printf("\nEnter fraction 1: numerator denominator:");
9.    scanf("%d%d", &num1, &deno1);
10.    printf("\nEnter fraction 2: numerator denominator:");
11.    scanf("%d%d", &num2, &deno2);
12. 
13.    Fract f1 = {num1, deno1};
14.    Fract f2 = {num2, deno2};
15.    Fract result = sum(f1, f2);
16.    printf("\nResult=%d/%d", result.num, result.deno);
17. 
18.    return 0;
19.}
20. 

//fract.h
typedef struct node {
    int num;
    int deno;
    }Fract;
Fract sum(Fract number1, Fract number2) {
    number1.num *= number2.deno;
    number2.num *= number1.deno;
    number1.deno *= number2.deno;
    number2.deno *= number1.deno;
    Fract sum;
    sum.deno = number1.deno;
    sum.num = number1.num + number2.num;
    return sum;
}

//标程
1.#ifndef Header_h
2.#define Header_h
3. 
4.typedef struct {
5.    int num;
6.    int deno;
7.} Fract;
8. 
9.Fract sum(Fract f1, Fract f2) {
10.    Fract result = {(f1.num * f2.deno) + (f2.num * f1.deno), f1.deno * f2.deno};
11.    return result;
12.}
13. 
14. 
15.#endif /* Header_h */

注意

函数的名字不能够直接进入赋值，必须要定义一个和函数一样的类型，返回函数值才能够，函数名称是用来调用的，不能够直接操作！