How can I find the number of elements in an array?

Question

I have an int array and I need to find the number of elements in it. I know it has something to do with sizeof but I\'m not sure how to use it exac

Answer 1

In real we can't count how many elements are store in array

But u can find the array length or size using sizeof operator.

But why we can't find how many elements are present in my array.

Because when we initialise a array compiler give memory on our program like a[10] (10 blocks of 4 size ) and every block has garbage value if we put some value in some index like a[0]=1,a[1]=2,a[3]=8; and other block has garbage value no one can tell which value Is garbage and which value is not garbage that's a reason we cannot calculate how many elements in a array. I hope this will help u to understand . Little concept

Answer 2

i mostly found a easy way to execute the length of array inside a loop just like that

 int array[] = {10, 20, 30, 40};
 int i;
 for (i = 0; i < array[i]; i++) {
    printf("%d\n", array[i]);
 }

Answer 3

I used following code as suggested above to evaluate number of elements in my 2-dimensional array:

#include <stdio.h>
#include <string.h>

void main(void)
{
    char strs[3][20] =
    {
        {"January"},
        {"February"},
        {""}
    };

    int arraysize = sizeof(strs)/sizeof(strs[0]);

    for (int i = 0; i < arraysize; i++)
    {
        printf("Month %d is: %s\n", i, strs[i]);
    }

}

It works nicely. As far as I know you can't mix up different data types in C arrays and also you should have the same size of all array elements (if I am right), therefore you can take advantage of that with this little trick:

1. count number of bytes with sizeof() function from whole 2d array (in this case 3*20 = 60 bytes)
2. count number of bytes with sizeof() function from first array element strs[0] (in this case 20 bytes)
3. divide whole size with size of one element what will give you number of elements

This snipped should be portable for 2d arrays in C however in other programming languages it could not work because you can use different data types within array with different sizes (like in JAVA).

Answer 4

void numel(int array1[100][100])
{
    int count=0;
    for(int i=0;i<100;i++)
    {
        for(int j=0;j<100;j++)
        {
            if(array1[i][j]!='\0') 
            {
                count++;
                //printf("\n%d-%d",array1[i][j],count);
            }
            else 
                break;
        }
    }
    printf("Number of elements=%d",count);
}
int main()
{   
    int r,arr[100][100]={0},c;
    printf("Enter the no. of rows: ");
    scanf("%d",&r);
    printf("\nEnter the no. of columns: ");
    scanf("%d",&c);
    printf("\nEnter the elements: ");
    for(int i=0;i<r;i++)
    {
        for(int j=0;j<c;j++)
        {
            scanf("%d",&arr[i][j]);
        }
    }
    numel(arr);
}

This shows the exact number of elements in matrix irrespective of the array size you mentioned while initilasing(IF that's what you meant)

Answer 5

If we don't know the number of elements in the array and when the input is given by the user at the run time. Then we can write the code as

C CODE:

while(scanf("%d",&array[count])==1) { 
  count++;
}

C++ CODE:

while(cin>>a[count]) {
  count++;
}

Now the count will be having the count of number of array elements which are entered.

Answer 6

I personally think that sizeof(a) / sizeof(*a) looks cleaner.

I also prefer to define it as a macro:

#define NUM(a) (sizeof(a) / sizeof(*a))

Then you can use it in for-loops, thusly:

for (i = 0; i < NUM(a); i++)