Choosing the greatest, lowest and middle number only with if/else

[Vichfret]

I made this program in C, as you may see I used a if for each case, my question is, is there a way to make this program in less lines using only if/else?



#include <stdio.h>
#include <stdlib.h>

main() {
int a, b, c;

system("clear"); // no windows allowed :P

printf("Introduce un número: \n"); // enter a number
scanf("%d", &a);
printf("Introduce un número: \n");
scanf("%d", &b);
printf("Introduce un número: \n");
scanf("%d", &c);

if(a == b || a == c || b == c) {
if(a == b && a == c && b == c)
printf("Los tres números son iguales\n"); // the three numbers are equal
else{
if(a == b)
if (a > c)
printf("%d es el mayor y %d es el menor\n", a, c); // x is the greatest and y is the lowest
else
printf("%d es el mayor y %d es el menor\n", c, a);
if(a == c)
if(a > b)
printf("%d es el mayor y %d es el menor\n", a, b);
else
printf("%d es el mayor y %d es el menor\n", b, a);
if(b == c)
if(b > a)
printf("%d es el mayor y %d es el menor\n", b, a);
else
printf("%d es el mayor y %d es el menor\n", a, b);
}
}

else {
if(a > b && a > c)
if(b > c)
printf("El numero mayor es %d, el de en medio es %d y el menor es %d \n", a, b, c); // the greatest number is x, the middle is y and the lowest is z
else
printf("El numero mayor es %d, el de en medio es %d y el menor es %d \n", a, c, b);
if(b > a && b > c)
if(a > c)
printf("El numero mayor es %d, el de en medio es %d y el menor es %d \n", b, a, c);
else
printf("El numero mayor es %d, el de en medio es %d y el menor es %d \n", b, c, a);
if(c > a && c > b)
if(a > b)
printf("El numero mayor es %d, el de en medio es %d y el menor es %d \n", c, a, b);
else
printf("El numero mayor es %d, el de en medio es %d y el menor es %d \n", c, b, a);
}

return 0;
}

[Mark20]

Try this link:

http://www.neowin.net/forum/lofiversion/index.php/t399541.html

[uljanow]

if a <= b:
if b <= c:
print "a, b, c"
elif a <= c:
print "a, c, b"
else:
print "c, a, b"
elif a <= c:
print "b, a, c"
elif b <= c:
print "b, c, a"
else:
print "c, b, a"

[Tony Flury]

with conditional operators :



#define GetMax(a,b) ((a) > (b) ? (a) : (b))
#define GetMax3(a,b,c) (max((a), max((b),(c))))
...
...
...

max = max3(x,y,z) ;
min = -(max3(-x,-y,-z)) :
...
...
...


The notice the min uses the max Macro deliberately - it is not a typo.

if you think about two numbers - say 99 and 1000 - clearly 1000 is bigger. now if you take the negative -99 and -1000 - clearly -99 is bigger which means that 99 (positive) must be the smaller of the two original numbers. It even works if the original numbers are a mix of postive and negative numbers.

You could of course define GetMax3 and GetMax as functions.

this wont satisfy the OP - who wanted only if/else - but that sounded like a home work problem anyway.

[Vichfret]

Nice, I only wanted to do my program different to my classmates' and I though I could do better with my logic since a if for each case doesn't sound like a good program

[Reiger]

Gah. This reminds me: given a function that chooses the largest can you use that function (and that function only apart from common numerical operations) to find the smallest?

The answer is yes. And the answer to that problem, for a 3 tuple also yields the answer to your problem. It is surprisingly elegant:

In Haskell:


module TupleTest where

sort3Tuple (a,b,c) = (h,m,l) -- return (highest, middle, lowest)
where
vals = [a,b,c] -- prepare values as a list
lMax = foldl max -- function to get the max out of a list and base-case value using a fold
h = lMax a vals -- get the maximum of the values in the tuple
inv = ((-1) *) -- inverter/negation function
l = inv (lMax (inv a) (map inv vals) ) -- get the minimum of the values in the tuple using the Max: this is done by negating values
m = (sum vals) - h - l -- get the remainder of substracting both maximum and minimum from the sum of all values in the tuple (== the remaining value not yet found)

[Lux Perpetua]

Yet another way:
int t;

if (b < a) t = a, a = b, b = t;
if (c < b) t = b, b = c, c = t;
if (b < a) t = a, a = b, b = t;

printf("%d, %d, %d\n", a, b, c);(untested, so bugs may reside, but hopefully the concept is clear).