Create an array of consecutive numbers from 1 to n in C?

Question

I'm trying to code a function that will make an array of consecutive numbers from 1 up to a number n . For example, I would expect array[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} . My code doesn't seem to be running but I don't exactly know why.

int count (int a)
{
   scanf("%d", &a);
   int i;
   int array[a];
   do
   {
      i = 0;
      a = a - i;
      array[a - 1] = a;
      i++;
   }
   while (a > 0);
   printf("Resulting array is %d", array[a]);
   return 0;
}

Answer 1

You should use a for loop to create your array.

int i;
int array[a];
for(i = 0; i < a; i++) {
  array[i] = i + 1;
}

return 0;

Answer 2

An idiomatic C way would be to define a function that fills an existing array:

void count(int *array, int max)
{
    for (int i = 0; i < max; ++i)
    {
        array[i] = i+1;
    }
}

You can call it eg like this:

int arr[10];
count(arr, 10);
for (int i = 0; i < 10; ++i)
{
    printf("%d ", arr[i]);
}

Answer 3

Your do while loop could simply be replaced by the following for loop:

for (i = 0; i < a; i++)
  array[i] = i + 1;

Similarly, for displaying each element in array , you can iterate over the elements of array by means of a for loop:

for (i = 0; i < a; i++) 
  printf("Element at %2d is %d\n", i, array[i]);

Answer 4

First and foremost, the print statement

 printf("Resulting array is %d", array[a]);

is wrong!! That's not how you print an array. You need to loop over the index, get the individual elements and print them.

Moreover, when you try using array[a] , think of the value for a . That's surely not what you want.

That said, there are other issues, like setting i to 0 in each iteration of do...while loop. That will lose the previously calculated value and start over. Move the initial assignment outside the loop body.

A slightly modified version, follow the comments

#include  <stdio.h>

int count (void)                 // no need for parameter if you're asking user for size
{
   int size = 0;                 //define size
   int ret = scanf("%d", &size); //read it

   if ( (ret != 1) || ( size < 1) ) return -1;      //error check

   int i = 0;                    //initialize local variables

   int array[size];              //note: this is VLA

   do
   {
       array[i] = i+1;           //start populating elements
       i++;
   }
   while (size > i);             //go on unless you've reached the given size

   puts ("Resulting array is");   // start printing ....

   for (i = 0; i < size; i++)
   {
       printf("%d\t", array[i]);   //...them one by one
   }

   return 0;
}

int main(void)
{
    (void)count();  //call the function
}

Answer 5

Try something like this:

int* count(int a)
{
   int array[a];
   for (int i = 0; i < a; ++i)
   {
      array[i] = i + 1;
   }
   for (int i = 0; i < a; ++i)
   {
      printf("%d, ", i + 1);
   }
   return array;
}

Answer 6

Just use a simple "for" loop:

for (i = 0; i < a; i++)
    {
    array [i] = i + 1;
    }

Answer 7

• You don't really need the a parameter. You can use a local
• the do,while is broken because you set i to 0 at each iteration and its overly complicated. You can achieve the same result in a much easier way. An element at location i has to have value i+1 , easy as it sounds.
• You have to loop over the element of the array to print them all. A single printf prints only 1 element of the array. If you want them all, then you have to iterate over all of them.

What about something like the following?

void count(){
   int a,i=0;
   scanf("%d", &a);
   int array[a];
   while(i<a){
    array[i]=i+1;
    i++
   }

   //now print it
   i=0;
   while(i<a)
    printf(" %d ", array[i++]);

}

---

As an aside, note that you are using VLAs. That means there's a limit on the size of the array you can create which depends on your machine+compiler.

Answer 8

So there are a few bugs here . One that if you are accepting 'a' as a arg to function , why are you scanning it then. It just over writes the value you collected. Two the way you are trying to print the array is wrong as specified above and you should use a loop. Three , you can save a few lines like ,

for(int i =0; i<a ; i++) 
{ 
arr[i]=i+1;
 }

Answer 9

I'm trying to code a function that will make an array of consecutive numbers from 1 up to a number n.

In C++ there is already such a function that is named iota and declared in the header <numeric> .

In C such a function can look the following way

void iota(int a[], size_t n, int value)
{
    for (size_t i = 0; i < n; i++) a[i] = value++;
}

Here is a demonstrative program

#include <stdio.h>

void iota(int a[], size_t n, int value)
{
    for (size_t i = 0; i < n; i++) a[i] = value++;
}

int main(void) 
{
    size_t n = 1;

    scanf( "%zu", &n );

    int a[n];

    iota( a, n, 1 );

    for ( size_t i =0; i < n; i++ ) printf( "%d ", a[i] );
    putchar( '\n' );

    return 0;
}

If to enter 10 then the output will look like

1 2 3 4 5 6 7 8 9 10

As for your program then in this loop

do
{
    i = 0;
    a = a - i;
    array[a - 1] = a;
    i++;
}
while (a > 0);

the variable a is not being changed because in the beginning of each iteration the variable i is set to 0.

You could write the loop like

i = 0;
do
{
    array[a - i - 1] = a - i;
    i++;
} while ( i != a );

Take into account that this statement

printf("Resulting array is %d", array[a]);

does not make sense. If you are going to output an array element by element you have to use a loop as for example it is shown in the demonstrative program above.