Making y[i] a modifiable variable in C

Question

I am building a program that randomly generates a password using the ascii tabe of values and can only contain one of each char. it generates a password that is 8 char long.

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

#define SIZE 10

char Charactor(char x);
void Check(char* y);

int main()
{
char string[SIZE];//defines varriables
char* point;
point=&string[SIZE];
srand(time(NULL));
for (int i=0;i<SIZE-1;i++)//empties out the string
{
    string[i]='\0';
}
for (int i=0;i<SIZE-2;i++)//randomizes a char for each space in the string
{
    string[i]=Charactor(*point);
}
Check(point);/checks the string for duplicated values
printf("%s\n",string);//prints string on screen
}

char Charactor(char x)
{
int rnd=0;
rnd=rand()%2;//chooses char or number using ascii
if (rnd==0)
{
    rnd=rand()%10+48;
}
else
{
    rnd=rand()%26+65;
}
x=(char)rnd;
return x;
}

void Check(char* y)
{
int run=0;
for (int i=0; i<SIZE-2;i++)
{
    for (int x=0; x<SIZE-2; x++) 
    {
        if (y[i]==y[x] && run=0)
        {
            run++;
            continue;
        }
        else
        {
            y[i]='\0';
            y[i]=Charactor(*y);
        }
    }
}
return;
}

with those changes the code is running now I just have to figure out how to change the correct value so I dont have any duplication.

Answer 1

Fix:

char* point =&string[0]; //Make it to point to first element

Since your Charactor(*point); is really not doing anything based on *point and later you use Check(point); probably to start a scan from start of string.

And

 if (y[i]==y[x] && run==0)
                      ^^Use Equality check

You cannot modify a boolean outcome of y[i]==y[x] && run as zero.

Note : However if (y[i]==y[x] && (run=0) ) wouldn't have thrown this error.

Answer 2

Your error seems to be that you are mistakenly setting run=0 in

if (y[i]==y[x] && run=0)

This is the part that most likely confuses your compiler. Doesn't have to do anything with Y.

Fix to:

if (y[i]==y[x] && run==0)