Newb Question concerning sizeof() function

I wrote a simple program that takes in 5 numbers into an array and then spits them back at you. Except that if the number inputted by the user is less than 10, it is doubled.

Simple enough. What I don't understand is why when I run the sizeof() function in my doubleA function, it returns the size of the array as 4. I assume it has something to do with its int value, but the sizeof() works fine in the main().

Halp?

Code:

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#include <stdio.h>

void doubleA(int a[]);

main()
{

        int array[5];
        int i;

        printf("Please enter 5 numbers: ");

        for(i=0; i<5; i++)
                scanf("%d", &array[i]);
       
        // My custom function
        doubleA(array);

        for(i=0; i<5; i++)
                printf("The %d number you printed was %d\n", i+1, array[i]);
        // prints sizeof() the array
        printf("size of array[] in main() function: %d\n", sizeof(array));
}

void doubleA(int a[])
{

        int i, arraysize;
        arraysize = sizeof(a)/sizeof(int);
        //prints sizeof() the array
        printf("size of a[] in doubleA = %d\n", sizeof(a));
        for (i=0; i<5; i++)
                if (a[i] <10)
                        a[i] *= 2;
}

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A couple of points.
1. sizeof is an operator (like + or *), not a function.

2. The method you're using to count the elements in the array only works when the actual array declaration is in scope. When you pass an array to a function, all you get is a pointer (so all the sizeof has to work on is a pointer).

The solution (like is used by many standard functions) is to also pass the array size as an additional parameter.

Ok, that makes sense. I just tested putting "sizeof(*a)" in the function to see if that helped but it didn't. :p

I guess I'll just accept the rule of thumb to pass the size as a parameter.

Cheers!

Quote:

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Originally Posted by Salem

A couple of points.
1. sizeof is an operator (like + or *), not a function.

2. The method you're using to count the elements in the array only works when the actual array declaration is in scope. When you pass an array to a function, all you get is a pointer (so all the sizeof has to work on is a pointer).

The solution (like is used by many standard functions) is to also pass the array size as an additional parameter.

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Remember sizeof is normally/always an compiler time operator; the value is a constant at run time.

Tim S.

Quote:

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Originally Posted by stahta01

Remember sizeof is normally/always an compiler time operator; the value is a constant at run time.

Tim S.

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Code:

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#include<stdio.h>
int main( void )
{
    int x;
    scanf( "%d", &x );
    if( x > 0 )
    {
        int array[ x ];
        printf( "array is %d\n", sizeof( array ) );
    }
    return 0;
}

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:trollface:

Quzah.

As always, thanks for keeping everyone honest quzah.:tongue:
VLAs have indeed introduced many interesting things to the mix when you look at compiler implementation, sizeof being one of those things. sizeof is still a unary compile time operator, however it's implementation for VLAs have required some trickery on the part of the compilers. I would be interested to see what solutions some of the makers have come up with.

I understand that the implementation of VLAs in general varies, with what it appears the common solution being simply placing a pointer to the VLA on the stack and then actually placing the array on the heap to avoid stack corruption. I imagine the solution for sizeof would be evaluating everything down to something along the lines of size*sizeof(underyling type) where size gets filled in at run time.

sizeof is compile time operator in C89. and in C99 can evaluate VLA size at run-time.
Edit:
As far as I understand, for int a[10] the number of element of part of the type. ie array of 10 integers.
Not sure what's the type of VLA?

Code:

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int n;
        int (*pa)[100];
        n = 11;
       
        int a[10][n];                    // no warning type check gone?
        // or  int a[10][101];      /// warning
        pa = a;

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Quote:

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Originally Posted by AndrewHunter

I imagine the solution for sizeof would be evaluating everything down to something along the lines of size*sizeof(underyling type) where size gets filled in at run time.

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Which is essentially the same way the size of any other array would be computed, except that the length is computed at compile time. Which sort of makes sense.

Quote:

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Originally Posted by Bayint Naung

Not sure what's the type of VLA?

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I seem to recall its type is "variable length array of <type> of the specified length" ;)