Finding middle value in C...

[DoEds]

Can anyone give me some ideas on how to find the middle value between 3 numbers.? I really dont know how to start...

I know how find minimum and maximum but finding the Middle value is kinda hard for me...

For those C Wizards give me some idea advice please?., just an idea u dont have to give codes... thx!

[rati]

Can anyone give me some ideas on how to find the middle value between 3 numbers.? I really dont know how to start...

I know how find minimum and maximum but finding the Middle value is kinda hard for me...

For those C Wizards give me some idea advice please?., just an idea u dont have to give codes... thx!

What exactly do you mean by middle value?
U can do it by using (a+b+c)/3
I hope its not that simple. Kindly explain what you mean by middle value

[thines01]

Do you mean something like this?:

#include <stdio.h>
#include <string.h>

int main(int argc, char** argv)
{
	char strData[128] = {'\0'};
	int intMiddle = 0;
	
	do
	{
		puts("Please enter some data");
		gets(strData);
	} while(NULL == strData[0]);
	
	
	intMiddle = (int)(strlen(strData)/2);
	
	printf("The middle is [%c]\n", strData[intMiddle]);
}

[Tom Gunn]

Can anyone give me some ideas on how to find the middle value between 3 numbers.?

Assuming it is always 3 numbers, and what you want is the median value, here is a simple way. Find the maximum then the minimum and the one you didn't pick is the median. :)

[mrnutty]

Also, if the numbers aren't too far apart, then what you could do
is get the average of the 3 numbers, and find the number that is closest
to the average. This will work long as the number aren't too far apart, like
1,2,100;

for example if the input was 1,2,3 and you know that the median is 2,
but if you found the average, totalSum/totalElement = (3+2+1)/3 = 6/3 = 2, then you see that this also gets the correct answer. Use it at your
own risk.

[DoEds]

#include <stdio.h> int main() { int a,b,c; int min,mid,max,i=1; printf("Enter number: "); scanf("%d",&a); printf("\nEnter number: "); scanf("%d",&b); printf("\nEnter number: "); scanf("%d",&c); if(((a<b)&&(b<c)) || ((a>b)&&(a<c))) {
        min=a;
        mid=b;
        max=c; } if(((b<a)&&(a>c)) || ((a>b)&&(a<c))) {
        min=c;
        mid=a;
        max=b; } if(((c<a)&&(c>b)) || ((c>a)&&(c<b))) {
        min=b;
        mid=c;
        max=a; } printf("\nMin is %d",min); printf("\nMid is %d",mid); printf("\nMax is %d",max); printf("\n"); for(i=min;i<=max;i++) {
        
        
        if(i==mid)
        continue;
        printf("\n%d",i); } getchar(); getchar(); }

Here's it is...I can find the min, mid and max...
But the problem is when i enter a Descending values like... a=3,b=2,c=1...
i got wrong output... try it,.,.

i dont know why? do i have a wrong if statement?

[jbennet]

To find out the media, use an array and a bubble sort to get them in order, then keep going until you are at the middle element.

[Iam3R]

I hope what he wants he is just " find the middle element when you have given three elements"

Ex: 1,3,6 the middle element is 3 ( this we cant get just by dividing)
5,6,7 the the middle element is 6.

if this is the case :

int find_mid(int n1, int n2, int n3) { int mid; if( n1 > n2 ) {
        if( n3 > n2)
        {
        if( n3 < n1 )
        mid = n3;
        else
        mid = n1;
        }
        else {
        mid = n2;
        } } else {
        if( n2 > n3 )
        {
        if ( n1 > n3)
        mid = n1;
        else
        mid = n3;
        }
        else{
        mid = n2;
        } } return mid ; }

the driver:

int main() {
        printf( "%d ", find_mid(3,4,7) );
        printf( "%d ", find_mid(5,2,8) );
        printf( "%d ", find_mid(2,9,4) );
        printf( "%d ", find_mid(7,5,9) );
        printf( "%d ", find_mid(7,4,6) );
        printf( "%d ", find_mid(1,3,6) );
        printf( "%d ", find_mid(5,6,7) ); return 0; }

[DoEds]

Thank you so much...

lam3r...

I really appreciate your help...

[krutarth]

hello frnds pls send me logic of fibo nacci series...........
my mai address is krutarthdoshi [e-mail]