第五阶段—函数—函数的传参—地址传递：从键盘输入一个值位n,分别求出1 + 3 + 5 +..+ n的奇数和。与0 + 2 + 4 + ...+ n -1的偶数和。

 1 #include<stdio.h>
 2 void sum(int num, int *x, int *y)
 3 {
 4     int flag = 0;
 5     int i = 0;
 6     *x = 0;
 7     for(i = 0;i <= num; i = i+2)//变量的递增条件时i= i+2
 8     {
 9         *x+=i;
10     }
11     
12     *y = 0;
13     for(i = 1;i <= num; i = i+2)
14     {
15         *y+=i;
16     }
17     return;
18 }
19 
20 int main(int argc, const char *argv[])
21 {
22     int n = 0;//定义需要传给形参，给形参赋值的变量
23     int even_sum = 0;
24     int old_sum = 0;
25     printf("plese input number:");
26     scanf("%d",&n);
27     
28     sum(n, &even_sum, &old_sum);
29     
30     printf("even_sum = %d,old_sum = %d\n",even_sum , old_sum);    
31     return 0;
32 }

标签：传参,函数,+..+,int,...+,void
来源： https://www.cnblogs.com/thismajor/p/15369108.html