error too few parameters to call even_odd

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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?

C++

# 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>

Posted 30-Sep-15 4:31am

Member 12014914

Updated 9-Jan-23 7:42am

Patrice T

v2

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Patrice T · Accepted Answer · 2015-09-30T04:44:00

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

C++

void even_odd(int n){

must be replaced by

C++

void even_odd()
    int n;{

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

C++

void even_odd();

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

C++

# 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>

Leo Chapiro · Accepted Answer · 2015-09-30T04:45:00

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

Your function

C

void even_odd(int n)

expects a parameter, but you call it without any:

even_odd();

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) ;  
}

Wendelius · Accepted Answer · 2015-09-30T04:51:00

Solution 3

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:

C++

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

Posted 30-Sep-15 4:51am

Wendelius

Comments

[no name] 30-Sep-15 10:51am

thank you guys

Leo Chapiro 30-Sep-15 10:54am

A good one, +5!

error too few parameters to call even_odd

3 solutions

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