Difference between assigning array directly and through function

Question

Say I have an existing array int *array

int *new_array = NULL;
new_array = array;

int *copy(int *array)
{
return array;
}

int* new_array = NULL;
new_array = copy(array);

What's the difference between the two?

Thanks for the help!

no difference except a function call, the result is exactly the same. Notice that you copied nothing, after operation both array and new_array points to the same address, meaning, the same data. — user8849929
– user8849929, Commented May 17, 2022 at 21:12
The two code snippets do the same thing, and in particular, neither one makes a copy of the array. — user3386109
– user3386109, Commented May 17, 2022 at 21:14
look at memcpy if you want to actually copy the data from array to new_array. — yano
– yano, Commented May 17, 2022 at 21:19
Does the actual code you are having a problem with look more like int *foo(void) { int array[something]; return array; } than int *copy(int *array) { return array; }? — Eric Postpischil
– Eric Postpischil, Commented May 17, 2022 at 21:40

0___________ · Accepted Answer · 2022-05-17 22:55:57Z

1

Arrays cannot be assigned in C language. In your examples, you assign a pointer with reference to the first element of the array. Your second snippet also only assigns pointer with pointer returned by the function. It does not copy data referenced by those pointers.

Say I have an existing array int *array

It is not an array - it is a pointer to int

If you want to copy elements of one array to another array you need to use function memcpy or memmove

int array1[5] = {1,2,3,4,5}
int array2[5];

memcpy(array2, array1, sizeof(array1));

answered May 17, 2022 at 22:55

0___________

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