Modifying array inside a function in C

Question

// test.txt
50
13
124

-

void hi ( int *b, FILE *pfile ) {

    rewind ( pfile );
    int i;
    for( i = 0 ; i < 3 ; i++ ) {
         fscanf ( pfile, "%d", &b[i] );
    }
}

int main ( void ) {

    FILE *fp = fopen ( "test.txt", "r" );
    int a[10];  //putting extra size for test.

    hi ( &a[0], fp );   
    printf("%d,%d,%d\n", a[0], a[1], a[2]);

    fclose ( fp );
    return 0;
}

I'm trying to understand pointers when there's array involved. While I was testing the code above, I noticed that by putting a different index value such as hi ( &a[0], fp ) to hi ( &a[1], fp ) I get different results.

//result of [ hi ( &a[0], fp ) ] //result of [ hi ( &a[1], fp ) ] 50,13,124 junk#,50,13 .

I am really confused about the results because in the 'hi' function I specify the start of the array from i = 0 which should mean it's storing the value starting from a[0]. But it seems like putting 1 instead of 0 is somehow moving values aside. Why is this happening?

Answer 1

You are not 'puuting a different size' but providing a pointer to the 1st element in the array and then the 2nd element.

Since in the second case you gave a pointer to the 2nd element, and tried to read 10 elements (starting at the 2nd position) but only have 9 valid elements when starting at the 2nd one, you are then reading into memory that is not in the array.

Answer 2

This might make more sense for you if you think of & as an operator which acts on a single item, much like - acts on a number to change the sign. In this case, & takes the address of whatever you give it. In the case of &a[0], you are taking the address of a[0]. In the case of &a[1], you take the address of a[1]. In other words, you take the address of different elements of the array. So in the function, b contains the address that you passed to it. As far as the function knows, what you pass in is the "first" element of the array. It knows nothing of any of the preceding elements.