取一个数字数组，然后只显示不同的数字

Question

In the below code I am taking an array to read 4 numbers and I need only to display the distinct number.

#include <stdio.h>

int main(){
    int ch[3];
    int s[3];
    int count = 0;
    int i;
    int j;

    for(i = 0; i < 4; i++){
        scanf("%d", &s[i]);
        for(i = 0; i < 4; i++){
            ch[i] = s[i];
            printf("ch= %d", ch[j]);
        }

        if(ch[i] == s[i]){
            count = count + 1;
            printf("%d =", count);
        }
    }
}

Somewhat I'm not getting the output, it is giving me a weird output.

Question:

What could be the cause of the strange output?

Answer 1

int ch[3];
int s[3];

This mean valid references to these arrays can be made from 0 to 2 , whereas your for loop is looping 4 times.

printf("ch= %d",ch[j]);

you mean ch[i] ?

what do you use "int j" for ?

Answer 2

Cant understand the question but i wrote a code for taking 4 inputs and giving the distinct number among them as an output(or the numbers which have only occurred once).

#include <stdio.h>
int main(){
   int ch[4];
   int s[4];
   int i;
   int j;
   for(j=0;j<4;j++){
    ch[j]=0;
    }

   for( i=0;i<4;i++){
    scanf("%d",&s[i]);
    }
   for(i=0;i<4;i++){
        for(j=0;j<4;j++){
            if(s[i]==s[j]){
                ch[i]++;
            }
        }
   }
   for(i=0;i<4;i++){
        if(ch[i]==1){
            printf("\n%d is the distinct number.", s[i]);
        }
   }
}