Accessing array via pointer in C syntax clarification

Question

This is a fairly basic question that I stumbled across today and want to know the difference between two syntaxes.

Let's say I have a function that assigns a value to an array that is pointed to. I noticed that while both syntaxes compile, the second one below seg faults and the first one runs fine. Why is this?:

Works fine:

foo(int** arr){
    for (i = 0; i < SUM; i++){
        (*arr)[i] = i+1;
    }
}

Seg fault:

foo(int** arr){
    for (i = 0; i < SUM; i++){
        *arr[i] = i+1;
    }
}

Example main:

main(){
    int* _arr;
    arr = (int *)malloc(sizeof(int)*50);//arbitrary
    foo(&_arr);
}

I wrote all this code as an example, if any clarification is needed let me know.

Why are you passing and using a double pointer in the function? a single pointer: void foo(int *arr) is enough. — David Ranieri
– David Ranieri, Commented Feb 21, 2017 at 21:02
could be useful if the malloc is done within the function. otherwise you're right, it's just food for segv. — Jean-François Fabre
– Jean-François Fabre ♦, Commented Feb 21, 2017 at 21:05

Jean-François Fabre · Accepted Answer · 2017-02-21 21:03:10Z

1

You're facing operator precedence / priority issues.

(*arr)[i] properly dereferences arr into an array, then adds i to get the value.
*arr[i] first takes arr+i (uninitialized memory if i>0: you have only 1 array) and tries to read from that invalid pointer: segfault

answered Feb 21, 2017 at 21:03

Jean-François Fabre♦

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2 Comments

Aroic Over a year ago

After I asked this question and got to thinking more about it that seemed like the answer, but I figured I'd post anyways just to make sure. Thanks!

ephemient Over a year ago

@TMartin To put it another way, *arr[i] has the same meaning as *(arr[i]). Sounds like you got it :)

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