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/*In an array of numbers, find how many times does
a number 'x' occurs and make a new array of them */

#include<stdio.h>

int main()
{
    int nums[] = {5,7,3,6,5,7,9,0,5,7};
    int repeat[10] = {0};
    int i,j,k, disqualify;
    int n[10]= {0};
    for( i=0; i < nums; i++)
    {
        k=0;
        n[i]=1;
        disqualify=0;

        for(j=i+1 ; j< nums; j++)
        {
            for(nums[i]; k <i ;k++)
            {
                if( nums[i] == repeat[k])
                {
                    disqualify =1;
                }
            }
            if( (nums[i] == nums[j]) && (disqualify=0))
            {
                repeat[i]= nums[i];
                n[i]++;
            }
        }
    }
    
    for(i=0 ; i< 10 ; i++)
    {
        printf("%d ", repeat[i]);
    }

    return 0;
}


What I have tried:

i think logic is right but the execution is not?
Posted
Updated 20-Oct-22 10:06am
v2
Comments
Member 15627495 20-Oct-22 14:57pm    
you got to fetch the array only two times. you do it 3 times.

one time as 'base' array, and a second time to check the redundant values.

n[i]++ can't work., because n is an array , the index is i. you can not increment an array like this.

instead of allocating the repeat number, in a loop , you have to count the occurs about the search number.

one loop inside a main loop will be enough, you are on a logical error. too much code.
merano99 20-Oct-22 15:50pm    
Of course, an instruction like n[i]++ would work and increment the contents of the array at location i.

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:
Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16
Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:
C
int Double(int value)
   {
   return value * value;
   }

Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.

This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!
 
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The two checks should not compile, because they make absolutely no sense.
C
for (i = 0; i < nums; i++)
for (j = i + 1; j< nums; j++)

If you use the length of the array instead of nums and assign value to the variable i, in the second loop j will be too large, because j = i + 1; definitely lies behind the array.

Another loop also contains nonsense code:
C
for (nums[i]; k <i; k++)

Possibly for(k = 0; k <i; k++) was meant?

I do not understand the meaning of "make a new array of them".

According to the program, characters that occur more than once are searched for, but then what should be in the new array?

Anyway, the condition
C
if ((nums[i] == nums[j]) &amp;&amp; (disqualify = 0))	{ ... }

is never fulfilled in this program and accordingly there are zeros in the result.
 
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v5
Comments
Member 15627495 20-Oct-22 16:47pm    
"make a new array of them" means to keep aside all the numbers existing more than one time.
If a number is two/three/four time in the original array, it must be save in another table, isolated.

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