Unexpected results while passing arrays as parameters in C?

Question

I am trying to get used to using pointers but i am struggling a bit whilde passing them into functions as parameters. I read that arrays are passed by reference to functions therefore a code like this can change the array elements(Assuming a valid array is given):

void f(int *b)
{
  b[2]=1;
}

in main:
f(validArray);

However even though we pass the pointer this doesn't seem to work;

void foo(int *b)
{
  b=b+2;
}

in main:
f(validArray);

doesn't move the pointer a step further. Why is that?

Vlad from Moscow · Accepted Answer · 2020-01-07 13:58:00Z

2

For starters arrays are non-modifiable lvalues. So you may not write for example

validArray = validArray + 2;

As for this pseudocode snippet

void foo(int *b)
{
    b=b+2;
}

in main:
foo(validArray);

then the pointer b is a local variable of the function. So within the function the value of b is changed due to this statement

    b=b+2;

But this does not have an effect on the passed argument.

Instead you could write

void foo( int **b )
{
    *b = *b + 2;
}

int main( void )
{
    int validArray[] = { 1, 2, 3, 4, 5 };

    int *p = validArray;
    foo( &p );
    printf( "%d\n", *p );
}

In C the term "passing by reference" means passing an object indirectly through a pointer to it. So dereferencing the pointer in a function you can change the pointed object. Pointers are the same objects. So if you want to change an original pointer you have to pass a pointer to the pointer.

edited Jan 7, 2020 at 13:58

answered Jan 7, 2020 at 13:55

Vlad from Moscow

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2 Comments

Jonaswg Over a year ago

f( &p ); should be foo( &p );

Vlad from Moscow Over a year ago

@Jonaswg Thanks. It was a typo.

Blaze · Accepted Answer · 2020-01-07 13:52:09Z

2

I read that arrays are passed by reference to functions

That's not quite right. When you pass an array to a function, a pointer is passed. Changing the pointer won't change the array. You can, however, change the array via the pointer. That's why the b[2]=1; works. b is still just a pointer that was passed as a copy though, so b=b+2; won't change the array.

It doesn't move the pointer a step further. Why is that?

It does move b ahead, but b is local to foo and is gone afterwards without changing the array. If you did this, for instance.

void foo(int *b)
{
    b=b+2;
    b[2]=1;
}

Then this will change validArray[4].

answered Jan 7, 2020 at 13:52

Blaze

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1 Comment

Mustafa Çığ Gökpınar Over a year ago

ohh because i am dereferencing the pointer b by using [] i can change the array indeed now i see.

Haris · Accepted Answer · 2020-01-07 13:54:11Z

2

b[2]=1;

is essentially

*(b+2) = 1;

which means its changing the data in the address (b+2)

b=b+2;

this means you are making b point to 2 elements next to what b was pointing before, which does not affect the data in the address of the pointer b, hence does not change the data in the array validArray

answered Jan 7, 2020 at 13:54

Haris

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