CODEWITHSUNDEEP
cpointersstruct
AlrzNmz
AlrzNmz

Reputation: 19

base operand of '->' has non-pointer value; but I'm almost certain it has pointer type?

I have defined a struct named wire:

typedef struct wire
{
    char type[9];
    char name[20];
    char value;
    struct io * inp=NULL;
    struct io * outp=NULL;
} wire;

and another named io:

typedef struct io
{
    struct io* next;
    struct log_gate* conn_gate ;
} io;

wires is a dynamic array of wire.

and compile error occurs in this line:

tmp=&wires[i]->outp;

The compile error is: base operand of '->' has non-pointer value 'wire'

Upvotes: 0

Views: 119

Answers (2)

Vlad from Moscow
Vlad from Moscow

Reputation: 311186

Write either

tmp = ( &wires[i] )->outp;

or

tmp = wires[i].outp;

If wires is a pointer to allocated memory of an array then wires[i] is an element of the array and has type struct wire that is it is not a pointer. So the correct syntax will be as the second example of statement shown above.

If you uses the address of this element like &wires[i] then unary operator & has lower priority than postfix operator ->. So you need to enclose expression &wires[i] in parentheses (&wires[i]) like it is shown in the first example of statement above that to get a pointer.

Upvotes: 2

Brian Tracy
Brian Tracy

Reputation: 6831

First we must take a look at the precedence of both operators in question. According to this chart, -> has a higher precedence than &. This means that we must evaluate the -> before the &. Lets take a look at the expression.

wires[i] /// this returns a wire

Due to the precedence of -> we now evaluate the struct accessor operation.

wires[i]->outp /// because wires[i] is a wire, not a wire *, the -> is misused.

wires[i] is just a plain old wire, not a pointer, so we must use the . operator to access its members. Remember, -> vs . is determined by the type of the struct, not the type of the member. Because the struct is not a pointer, we use .

The precedence issue is important because if & had a higher precendence than ->, then the expression &wires[i] would be evaluated first. This would mean that we are trying to get at the outp field of a wire * instead of a wire, and the -> operator would be justified.

Upvotes: 2

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