c语言 函数的参数传递示例
Define a function with no argument and no return type in C language.
用C语言定义一个没有参数且没有返回类型的函数。
In the program, we have function named fun1 which has no argument and no return type (void is the return type - that means, function will not return anything).
在程序中，我们有一个名为fun1的函数，它没有参数，也没有返回类型( void是返回类型-这意味着该函数将不返回任何内容)。
Within the function fun1 we are declaring an array, calculating sum of its elements and printing the sum.
在fun1函数中，我们声明一个数组，计算其元素的总和并打印总和。
程序 (Program)
</ s> </ s> </ s>/*
User define function example with no argument
and no return type
*/
#include <stdio.h>
void fun1(void)
{
int array[10]={1,2,3,4,5,6};
int i=0,sum=0;
for(i=0;i<6;i++)
{
sum = sum + array[i];
}
printf("\nThe sum of all array elements is : %d",sum);
}
// main function
int main()
{
//calling the function
fun1();
return 0;
}
Output
输出量
The sum of all array elements is : 21
翻译自: https://www.includehelp.com/c-programs/user-define-function-example-with-no-argument-and-no-return-type.aspx
c语言 函数的参数传递示例
