Using sizeof function in c

Question

I want to write a function in c which is able to print the array we created. However, there is a problem. When I want to create the loop I need to know the number of elements in the array. So, I used the sizeof function in order to fix that.

I wrote the line size = sizeof(list) / sizeof(list[0]); in order to get the size of the whole array and divide it to size of one element in order to get the number of elements in the array. However, when I build the code I get the following warning:

warning : 'sizeof' on array function parameter 'list' will return the size of 'int *'

#include <stdio.h>

void displayArray(int myarray[]);

int main()
{
    int list1[] = {0, 5, 3, 5, 7, 9, 5};
    displayArray(list1);
    return 0;
}

void displayArray(int list[])
{
    int size;
    size = sizeof(list) / sizeof(list[0]);
    printf("Your array is: ");
    for (int i = 0; i < size; i++)
    {
         printf("%d ", list[i]);
    }
    printf("\n");
    return;
}

Answer 1

void displayArray(int list[]) is exactly the same as void displayArray(int *list). This is because an array doesn't have a value itself, and when you try to evaluate just the identifier of the array, what you get instead is a pointer to its first element.

Therefore, passing an array to a function always means passing a pointer. The sizeof operator (it's not a function) doesn't get your array type but a pointer type, and the size of the pointer is useless to you.

A typical approach is to explicitly pass the size of the array as a second parameter to your function, the correct type for this is size_t.

Answer 2

There's no such thing as array parameters. They're just hidden pointers.

You should size your array in the caller (where there's a real array) and pass the size to the callee.

(As arrays get passed to callees, their size information gets lots in the array-to-pointer conversion, and so you have to pass it separately)

#include <stdio.h>

void displayArray(int myarray[], int size);

int main()
{
    int list1[] = {0, 5, 3, 5, 7, 9, 5};
    displayArray(list1, sizeof list1 / sizeof list1[0]);
    /*you don't need the parens if the argument is a variable / isn't a type */
    return 0;
}

void displayArray(int list[], int size)
{
    printf("Your array is: ");
    for (int i = 0; i < size; i++)
    {
         printf("%d ", list[i]);
    }
    printf("\n");
    return;
}

Answer 3

According to the C Standard (6.3.2.1 Lvalues, arrays, and function designators)

3 Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue. If the array object has register storage class, the behavior is undefined.

And (6.7.6.3 Function declarators (including prototypes))

7 A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation...

Thus the following function declarations declare the same one function

void displayArray(int myarray[100]);
void displayArray(int myarray[10]);
void displayArray(int myarray[1]);
void displayArray(int myarray[]);
void displayArray(int *myarray);

You may include all these declarations (that are not definitions) in your program. The function parameter that declares an array is adjusted to pointer type.

So you need explicitly pass to the function the size of the array if the array does not have a sentinel value. For example

void displayArray( const int *myarray, size_t n );

//...

displayArray( list1, sizeof( list1 ) / sizeof( *list1 ) );

Answer 4

You should have create a variable to contain the size of the array and add this as a parameter to the function. 'list' is just a pointer to the first element of the array, so when you check sizeof(list), it returns the size of the pointer. Try it like this:

#include <stdio.h>
void displayArray(int myarray[]);
int main()
{
    int list1[] = {0, 5, 3, 5, 7, 9, 5};
    int size1 = 7;
    displayArray(list1,size1);
    return 0;
}

void displayArray(int list[], int size)
{
    printf("Your array is: ");
    for (int i = 0; i < size; i++)
    {
         printf("%d ", list[i]);
    }
    printf("\n");
    return;
}