Question

Possible Duplicate:
Sizeof an array in the C programming language?

Why is the size of my int array changing when passed into a function?

I have this in my main:

int numbers[1];
numbers[0] = 1;
printf("numbers size %i", sizeof(numbers));
printSize(numbers);
return 0;

and this is the printSize method

void printSize(int numbers[]){
printf("numbers size %i", sizeof(numbers));}

You can see that I dont do anything else to the numbers array but the size changes when it gets to the printSize method…? If I use the value of *numbers it prints the right size…?

Answer 1

Any array argument to a function will decay to a pointer to the first element of the array. So in actual fact, your function void printSize(int[]) effectively has the signature void printSize(int*). In full, it’s equivalent to:

void printSize(int * numbers)
{
    printf("numbers size %i", sizeof(numbers));
}

Writing this way hopefully makes it a bit clearer that you are looking at the size of a pointer, and not the original array.

As usual, I recommend the C book’s explanation of this 🙂