错误：“c中未分配指针”

Question

this error is always fired, when i'm try to free my allocated struct the second time, which it shouldn't, because the struct is set to NULL after i'm freeing it.

here's my struct with no real pointer inside it:

 typedef struct{
        int frame; 
        double timestamp; 
        int identifier; 
        int state; 
        int unknown1; 
        int unknown2; 
        mtVector normalized;
        float size; 
        int unknown3;
        float angle; 
        float majorAxis;
        float minorAxis;
        mtVector unknown4; 
        int unknown5[2]; 
        float unknown6; 
    }Touch;

the barebone main function:

int main(){
    Touch *myTouch = NULL;
    int inputCounter = 0;
    //whenever  a touch is recognized:
    ...
    myTouch = (Touch*)realloc(myTouch,sizeof(Touch)*(inputCounter++));
    ...
    // everything works fine until:
    freeTouch(myTouch);
}

void freeTouch(Touch *f){
    if(f != NULL){
        free(f);
        f = NULL;
    }
}

anybody got an idea?

Answer 1

f is a local variable. free(f) will affect the allocated memory, but f = NULL has no impact on myTouch in freeTouch(myTouch); .

Try

void freeTouch(Touch **f){
    if(*f != NULL){
        free(*f);
        *f = NULL;
    }
}

instead and use freeTouch(&myTouch) .

Answer 2

You have two problems there. The first is that it's not a good idea to explicitly cast the return value from malloc or realloc . Doing so can cause problems if you forget to include the prototype/header for it.

Secondly, freeing f within the function frees the local copy. Until C gains references, there are two possibilities. First pass a pointer to the pointer and use that:

void freeTouch (Touch **pF){
    if (*pF != NULL){
        free (*pF);
        *pF = NULL;
    }
}
:
freeTouch (&myTouch);

or pass back NULL so you can assign:

void *freeTouch (Touch *f){
    free (f);
    return NULL;
}
:
myTouch = freeTouch (myTouch);

You'll notice that the second one doesn't care whether you pass in NULL - it's perfectly acceptable to try an free the NULL pointer since it's effectively a no-op (other than the function call itself).

Answer 3

First of all, never use

x = realloc(x, size);

because if x is allocated before and realloc fails, you make it NULL while the memory is still there and therefore you create garbage.

Second,

void freeTouch(Touch *f);

gets a pointer by value and therefore cannot change the pointer itself. So your f = NULL; is not effective. You need to change your code to:

int main(){
    Touch *myTouch = NULL, temp;
    int inputCounter = 0;
    //whenever  a touch is recognized:
    ...
    temp = realloc(myTouch,sizeof(*temp) * (inputCounter++));
    if (temp == NULL)
        /* handle error */
    myTouch = temp;
    ...
    // everything works fine until:
    freeTouch(&myTouch);
}

void freeTouch(Touch **f){
    if(f != NULL && *f != NULL){
        free(*f);
        *f = NULL;
    }
}

Sidenote: It's a good idea to use realloc (and likewise malloc ) like this:

x = realloc(count * sizeof(*x));

There is no need to cast the output or realloc . Also, sizeof(*x) allows you to not repeat the type of x every time.