使用free（）时的分段错误

Question

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?

Answer 1

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.}

Answer 2

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.

Answer 3

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().

Answer 4

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

Answer 5

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.