错误太少的参数调用even_odd [英] error too few parameters to call even_odd

问题描述

我想显示从1到给定数字的奇数和偶数整数，并显示奇数和和的总和，但总是出现错误，说的参数太少。 （无效功能是必需的）最后我应该把公式用于获得奇数和总和的公式？

Hi, I would like to display the odd and even integers from 1 upto given number and display the sum of odd and sum of even but there's always an error saying too few parameters. (void function is required) And lastly where should i put the formula for getting the sum of odd and sum of even?

# include <stdio.h>  
# include <conio.h>  

void even_odd(int n);

int main()  
{  
    int n, i ;  
    clrscr() ;  
    even_odd();
    getch() ;  
    return 0;
}  

void even_odd(int n){
    printf("Enter Integer : ") ;  
    scanf("%d", &n) ;  
    printf("\nThe odd numbers are :\n\n") ;  
    for(i = 1 ; i <= n ; i = i + 2)  
        printf("%d\t", i) ;  
    printf("\n\nThe even numbers are :\n\n") ;  
    for(i = 2 ; i <= n ; i = i + 2)  
        printf("%d\t", i) ;  
}
</conio.h></stdio.h>

推荐答案

你的问题是 n 不是参数，而是局部变量

Your problem is that n is not a parameter but a local variable

void even_odd(int n){

必须替换为

must be replaced by

void even_odd()
    int n;{

哎呀，您还需要在 main 之前将行更改为

Ooops, you also need to change the line before main to

void even_odd();

如果您按原样编译，Turbo C会告诉您还有另一个问题，因为 i 和 n 用于 even_odd ，但在 main 中声明。

两次更正的源代码看起来像

If you compile as is, Turbo C will tell you that there is another problem because i and n are used in even_odd but declared in main.
the source code with both corrections will look like

# include <stdio.h>
# include <conio.h>

void even_odd();

int main()
{
    clrscr() ;
    even_odd();
    getch() ;
    return 0;
}

void even_odd(){
    int n, i ;
    printf("Enter Integer : ") ;
    scanf("%d", &n) ;
    printf("\nThe odd numbers are :\n\n") ;
    for(i = 1 ; i <= n ; i = i + 2)
        printf("%d\t", i) ;
    printf("\n\nThe even numbers are :\n\n") ;
    for(i = 2 ; i <= n ; i = i + 2)
        printf("%d\t", i) ;
}
</conio.h></stdio.h>

引用：

错误说参数太少。 （需要空函数）

an error saying too few parameters. (void function is required)

你的函数

Your function

void even_odd(int n) 

需要一个参数，但你不用任何参数调用它：

expects a parameter, but you call it without any:

even_odd();

试试这样：

Try it like this:

# include <stdio.h>  
# include <conio.h>  
 
void even_odd();
 
int main()  
{  
    int i ;  
    clrscr();
    even_odd();
    getch() ;
    return 0;
}  
 
void even_odd()
{
    int i, n;
    printf("Enter Integer : ") ;  
    scanf("%d", &n) ; 
	
    printf("\nThe odd numbers are :\n\n") ;  
    for(i = 1 ; i <= n ; i = i + 2)  
        printf("%d\t", i) ;

    printf("\n\nThe even numbers are :\n\n") ;  
    for(i = 2 ; i <= n ; i = i + 2)  
        printf("%d\t", i) ;  
}

其他人已经描述了调用的问题。那么计算总和又是什么呢？你已经有两个循环，一个用于奇数，另一个用于偶数，为什么不计算这些循环中的总和。



类似于：

Others have already described the problem with the call. So what comes to calculating the sum; You already have two loops, one for odd and another for even numbers, why not calculate the sum inside those loops.

Something like:

...
for(i = 1 ; i <= n ; i = i + 2) {
   printf("%d\t", i) ;  
   odd_sum = odd_sum + i;
}
...

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