Why can't access struct pointer outside of function

Question

Describe Issue: I'm able to access buffer variable inside of malloc function and can retrieve and set data with no issues

any attempt to access *(buffer+ insert some index here)->data outside of malloc function results in following error

mem* demo = malloc(2);
if(*(demo+1)->data == 0x00) {
... do some stuff here 
}

following error is produced by gcc cross compiler

kernel.c:96:21: error: invalid type argument of '->' (have 'int')
   96 |         if(*(demo+1)->data == 0x00) {

Code:

//Licensed under public domain
//also please note there is no standart library this is on a embedded system

typedef struct{
    _Bool allocated;
    unsigned char data;
} mem;
mem memory[1000];

mem* malloc(size_t size){
    mem* buffer[size];
    unsigned int successfulCounts = 0;
    unsigned int bufferCounter = 0;
    for(unsigned int i = 0; i < sizeof(memory); i++){
        //Hey that's available memory for us
        if(memory[i].allocated == 0){
            //because buffer is 16 4 items in memory (16*4)-15*4 can be found like this
            if(successfulCounts < sizeof(buffer)-sizeof(buffer-1)){
                *(buffer+successfulCounts) = &memory[i];
                successfulCounts++;
                memory[i].allocated = 1;
            }else{
                break;
            }
        }
    }
    return buffer;    
}


//... some more code that implements stuff like free() and calloc()

Odd Findings:

when mem * in function changed to unsigned char and returned *(buffer+1) i can access the data for some odd reason and i can get the exact same data i have pushed nothing is corrupted as i expect for some odd reason

@klutt example has been fixed now gcc in.c -ffreestanding -nostdlib -o test.elf needs a int main() though — lplplplp
– lplplplp, Commented Jul 15, 2022 at 22:27

Vlad from Moscow · Accepted Answer · 2022-07-15 22:53:02Z

0

This if statement

if(*(demo+1)->data == 0x00) {

is equivalent to

if( *( ( demo + 1 )->data ) == 0x00) {

but data is not a pointer. It has the type unsigned char

typedef struct{
    _Bool allocated;
    unsigned char data;
} mem;

It seems you mean

if( (demo+1)->data == 0x00) {

Pay attention to that the function in any case is invalid

mem* malloc(size_t size){
    mem* buffer[size];
    //...
    return buffer;    
}

For starters the return type of the function is mem * while the type of the returned expression is mem **. And moreover the function returns a pointer to a local object (array). So the returned pointer will not be invalid because the array will not be alive after exiting the function.

edited Jul 15, 2022 at 22:53

answered Jul 15, 2022 at 22:11

Vlad from Moscow

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5 Comments

lplplplp Over a year ago

still the same issue kernel.c:96:20: error: invalid type argument of '->' (have 'int') 96 | if((demo+1)->data == 0x00) {

Vlad from Moscow Over a year ago

@lplplplp The expression demo + 1 does not have the type int. it is a pointer provided that it is defined like mem* demo = malloc(2);

Chris Dodd Over a year ago

No it is not -- -> is higher precedence than *, so it is equivalent to if (*((demo+1)->data) == 0x00)

Chris Dodd Over a year ago

From the error message provided, demo has type int (as does demo + 1), so the demo in the statement with the error is not the demo in the earlier declaration. That's why an MRE is so important.

Vlad from Moscow Over a year ago

@ChrisDodd You are right. I was not attentive. But now I can not remove the answer because it is marked as the best.:)

Maxwell D. Dorliea · Accepted Answer · 2022-07-15 22:40:41Z

0

Your problem is you are de-referencing the pointer before using the arrow operator (->).

The arrow operator(->) is used when you want to access a member of a struct using a pointer.

struct new_t
{
    char *str;
    int num;
};

struct new_t p1 = {"Dog", 5};
struct *ptr = &p;

char *s = ptr->str; //is valid

char *s1 = ptr.str; // is not valid

char *s2 = *(ptr)->str; // is not valid

Whereas the dot(.) operator is used to access a member of a struct

int n = p1.num; //is valid

n = p1->num; //is not valid

To solved your problem used this

if(*(demo+1).data == 0x00) {
... do some stuff here 
}

or

if((demo+1)->data == 0x00) {
... do some stuff here 
}

answered Jul 15, 2022 at 22:40

Maxwell D. Dorliea

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