CODEWITHSUNDEEP
c++cstring
learner
learner

Reputation: 335

segmentation fault when free() is used

This code causes a Segmentation Fault:

int main(){

    char *p;
    char a[50] = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
    p = (char *)malloc(50*sizeof(char));

    if(!p){
            cout << "Allocation Failure";
            cout << "\n";
    }
    else{
            cout << "Allocation Success";
            cout << "\n";
            p = a;
            cout << p;
            cout << "\n";
            free(p);
    }

    return 0;
}

The output after executing this program is:

Allocation Success

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Segmentation fault

I am not able to find the bug. What may be reason?

Upvotes: 4

Views: 2251

Answers (5)

pb2q
pb2q

Reputation: 59627

You're calling free on a block of memory that wasn't allocated using one of the malloc methods.

When you do: p = a you're assigning to p the memory used by the stack array a. That memory wasn't allocated using malloc and hence it can't be freed using free.

Furthermore with that re-assignment, you'll lose track of the block that you originally allocated with malloc and assigned to p, causing a memory leak.

Upvotes: 10

mohit
mohit

Reputation: 6064

p = a; // p points a

Executing this line, p should be pointing to const string (char *). You cannot free() anything that is not obtained by calling malloc or calloc. Since in this example you try to free() a const string, you get an error.

Upvotes: 0

pizza
pizza

Reputation: 7640

you actually meant memcpy(p,a,50); not p=a, remember C does not have a string data type.

Upvotes: 0

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272657

This:

p = a;

copies the pointer, not the contents of the pointed-to memory. p now points at the first element of the array a. So when you do free(p), you're trying to free a non-dynamic array, which doesn't make any sense.1

You should investigate strncpy() to copy strings.

1. And it also causes a memory leak.

Upvotes: 12

Crashworks
Crashworks

Reputation: 41444

char a[50] allocates an array of fifty characters on the stack. a therefore points at an address on the stack.

p = a sets p to point at the same address as a. Therefore after p = a, p points at an address on the stack. Remember, p contains an address.

Afterwards, free(p) tries to free a region of memory on the stack, which is illegal. You can only free memory you got from malloc().

Upvotes: 0

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