invalid type argument of ‘->’

Question

i am getting error as "invalid type argument of ‘->’ " on the below two marked lines please suggest how to correct it

#include<stdio.h>

struct arr{
    int distance;
    int vertex;
};

struct heap{
    struct arr * array;

     int count; //# of elements
     int capacity;// size of heap
     int heapType; // min heap or max heap
};


int main(){
    int i;
    struct heap * H=(struct heap *)malloc(sizeof(struct heap));
    H->array=(struct arr *)malloc(10*sizeof(struct arr));

    H->array[0]->distance=20;//error

    i=H->array[0]->distance;//error

    printf("%d",i);
}

`H->array` contains 10 `struct arr`, not 10 `struct arr *`. So there's no need for the extra dereferencing. The first line should be `H->array[0].distance = 20;` — Praetorian, Sep 17 '13 at 21:24
[You do not need to cast the result of malloc.](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — pzaenger, Sep 17 '13 at 21:25
@Praetorian - Why are you answering the question in a comment and not as an actual answer? — Andrew Cottrell, Sep 17 '13 at 21:29
@pzaenger I'd better say: **DO NOT** cast the result of malloc. — LorenzoDonati4Ukraine-OnStrike, Sep 17 '13 at 21:51
@Andrew To gain comment rep, of course! :P I didn't think that terse comment was worthy of an answer, and didn't have time to type a more detailed answer. — Praetorian, Sep 17 '13 at 21:58

score 7 · Accepted Answer · answered Sep 17 '13 at 21:24

The left argument of -> must be a pointer. H->array[0] is a structure, not a pointer to a structure. So you should use the . operator to access a member:

H->array[0].distance = 20;
i = H->array[0].distance;

or combine them:

i = H->array[0].distance = 20;

BTW, in C you should not cast the result of malloc(). malloc() returns void*, and C automatically coerces this to the target type. If you forget to #include the declaration of malloc(), the cast will suppress the warning you should get. This isn't true in C++, but you usually should prefer new rather than malloc() in C++.

score 1 · Answer 2 · answered Sep 17 '13 at 21:39

1

The subscript implicitly dereferences the array member. If array has type struct arr *, then array[0] has type struct arr (remember that a[i] is equivalent to *(a + i));

answered Sep 17 '13 at 21:39

John Bode

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invalid type argument of ‘->’

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