C While loop for an array of integers

Question

This is a noob question. I have an array called Counter[N][N] and I want to do something like:

While (each element of Counter < 10000) {do something}

While (there exists an element of Counter < 10000) {do something}

Is there an easy way of doing that in C?

Answer 1

Easy if you can write a function which will return 1 if all the elements are less than 10000 :

int check_array_lt(int row_count, int col_count, int** array, int value)
{
  int i,j;
  for(i=0;i<row_count;i++)
    for(j=0;j<row_count;j++)
      if (array[i][j]>=value)
        return 0;
  return 1;
}

then use it :

while( check_array_lt(N,N,counter,10000) ) {
  do something
}

---

For the second version of the question (no more 'each element < 10000' but 'at least one element < 10000') :

int check_array_lt_atleast(int row_count, int col_count, int** array, int value)
{
  int i,j;
  for(i=0;i<row_count;i++)
    for(j=0;j<row_count;j++)
      if (array[i][j]<value)
        return 1;
  return 0;
}

---

As stated by Jonathan Leffler, this solution works only if the array is dynamically created; if Counter is declared as an array with #defined N, than my solution decays in Jonathan's one.

Answer 2

What about this?

 int flag=0;
        for(i=o;i<n;i++)
        {
            for(j=o;j<n;j++)
            {
                if(counter[i][j]<10000)
                    //statements;
                else
                {
                  flag=1;
                    break;
                }

            }
  if(flag==1)
    break;
        }

Answer 3

This function tests whether the counter array passed in has an element smaller than the specified value:

bool has_element_less_than(int value, int counter[N][N])
{
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
             if (counter[i][j] < value)
                 return true;
        }
    }

    return false;
}

You use it:

if (has_element_less_than(10000, counter))
    do_something();

You could deal with variable dimension arrays in C99 by passing N as a parameter to the function. It assumes you have the C99 header <stdbool.h> available and included.

---

Is this what you're after? You mention 'While' so it isn't clear whether you need to use a while loop — if you do, I think this does the job:

int i = 0;

while (i < N)
{
    int j = 0;
    while (j < N)
    {
         if (counter[i][j] < 10000)
         {
             counter[i][j] = do_something(i, j, counter[i][j]);
         }
         j++;
    }
    i++;
}

Or, more colloquially but using for loops:

for (int i = 0; i < N; i++)
{
    for (int j = 0; j < N; j++)
    {
         if (counter[i][j] < 10000)
         {
             counter[i][j] = do_something(i, j, counter[i][j]);
         }
    }
}

Note that this code is using C99; you can declare i and j outside the loops and it becomes C89 code. Also, if for any reason you need i or j (or, more likely, both) after the loop, you need to declare the variables outside the loop.

The second solution with for loops is more idiomatic C; the for loop is very good for this job and is what you should plan to use, not least because it packages all the loop controls on a single line, unlike the while loop solution which has the initialize code on one line (outside the loop), the condition on another, and the reinitialization on yet a third line.

Answer 4

this can be nicely done with pointers

  while(true)
  {
     int* pEnd = &Counter[0][0] + N*N;
     int* pCurrent = &Counter[0][0];

     bool AnyLess = false;
     while(pCurrent < pEnd && !AnyLess) { AnyLess |= *pCurrent++ < 10000; } 
     if(!AnyLess)
        break;

  }

Answer 5

although you did not ask for c#, here is the c# solution because it would be so easy:

    var Counter = new int[N,N];
    while(Counter.Cast<int>.Any(i => i < 10000)) { dosomething(); } 

Answer 6

Solution for EDITED question

    for(i=0;i<N;i++)
    for(j=0;j<N;j++){

    if(counter[i][j]<10000)
    {
       //Do something
    }else
       goto OUT;
    }

    OUT:
    printf("out of loops");

In my arrogant opinion

goto

statement is the most elegant construct in programming language . If anybody wants to become billionaire like Bill Gates , they must use lots and lots of goto's while developing their path breaking software .

Answer 7

You could do

for(int x = 0; x < N; x++) {
    for(int y = 0; y < N; y++) {
        if (Counter[x][y] < 10000){
             //Do something with Counter[x][y]
        }
    }
}

Answer 8

Use nested for loop

for(i=0;i<N;i++)
for(j=0;j<N;j++){

if(counter[i][j]<10000)
{
//Do something
}
}