Biggest and smallest of four integers (No arrays, no functions, fewest 'if' statements)

Question

You see, I\'ve self-taught myself C++ (not completely, I\'m still procrastinating -_-). So, now I started university and they\'re teaching C and they made us do a program of

Answer 1

This is C code has only 4 if statements. It moves max number to d position and min number to a position. Values b and c are not properly arranged within a sequence, but since requirements ask for min and max this code completes a job:

#include <stdio.h>


    int main() {
        int a, b, c, d, temp;
        printf("Enter four digits: ");
        scanf("%d %d %d %d", &a, &b, &c, &d);
        if ( a > b){
            temp = a; a = b ; b = temp;
        }
        if ( c > d){
            temp = c; c = d ; d = temp;
        }
        if ( b > d ){
            temp = b; b = d; d = temp;
        }
        if ( a > c){
            temp = a; a = c ; c = temp;
        }
        printf("Max %d\nMin %d\n", d, a);

        return 0;
    }

Answer 2

I saw this answer that used for loop and if else decision statement , however I will post a solution that I suppose will run faster and will use only four variables. So here it goes...

#include<stdio.h>
void main()
{
int a,b,c,d;
printf("Enter four numbers of your choice");
scanf("%d%d%d%d",&a,&b,&c,&d);
a>b&&a>c?a>d?printf("%d",a):printf("%d" ,d):(b>c&&b>d)?printf("%d",b):c>d?printf("%d", c):printf("%d",d);
}

Answer 3

Try something like this

int main(void) {
    int a=-2,b=-3,c=-4,d=-5;
    int max=a,min=a;

    if(b>max){
        max=b;
    }else if(b<min){
        min=b;
    }
    if(c>max){
        max=c;
    }else if(c<min){
        min=c;
    }
    if(d>max){
        max=d;
    }else if(d<min){
        min=d;
    }
    printf("max: %d min : %d",max,min);
    return 0;
}

Demo

Answer 4

#include <stdio.h>

int max_of_four(int a, int b, int c, int d){
    int mx_A_B = (a > b) * a + (a <= b) * b;
    int mx_C_D = (c > d) * c + (c <= d) * d;

    return (mx_A_B > mx_C_D) * mx_A_B + (mx_A_B <= mx_C_D) * mx_C_D;
}


int main() {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int ans = max_of_four(a, b, c, d);
    printf("%d", ans);
    
    return 0;
}

Answer 5

This doesn't use any loops and only 4 conditionals:

#include <stdio.h>

int main(void)
{
  int i, j, k, l, t, max, min;

  printf("Enter four integers: ");
  scanf("%d%d%d%d", &i, &j, &k, &l);

  if (i < j) {
    t = i;
    i = j;
    j = t;
  }
  if (k < l) {
    t = k;
    k = l;
    l = t;
  }
  max = (i >= k) ? i : k;
  min = (j <= l) ? j : l;

  printf("Largest: %d\n", max);
  printf("Smallest: %d", min);

  return 0;
}
  

  

Answer 6

Here is a solution with no if or elseif or function or macro, but using bit-shift and subtraction instead; only using a single for loop:

#include <stdio.h>
int main(){

  int num , max, min;

  printf("Enter four numbers: ");
  scanf("%d", &num);
  max = min = num;

  for(int i = 0; i < 3; i++)
  { 
    scanf("%d", &num);
    max = max * (1 - ( (max-num) >> 31) )
        + num *      ( (max-num) >> 31);
    min = min * (1 - ( (num-min) >> 31) )
        + num *      ( (num-min) >> 31);
  }

  printf("\n%d %d", max, min);
  return 0;
}

The (max-num) >> 31) operation captures the sign of the difference, which when multiplied by the second number yields the minimum value of the comparison.

This comes from an old SQL coding trick from the days before there was a CASE WHEN construct in that language.