I wrote a code expecting to receive error along with the description expression must be a modifiable value but i didn't, I don't understand can arrays that were dynamically allocated be modified?
{
int* x;
x =(int*) malloc(3 * sizeof(int));
if (x == 0)
{
printf("sorry met an error");
exit(1);
}
x[0] = 0;
x[1] = 1;
x[2] = 2;
printf("%p\n", x);
printf("%d\n", x[0]);
printf("%d\n", sizeof(x));
x++;
printf("%p\n", x);
printf("%d\n", x[0]);
printf("%d\n", x[1]);
printf("%d", sizeof(x));
free(x);
return 0;
}
By the way the free function here is also triggering a breakpoint any ideas why?
x is a pointer to the first element in an array. (x is not an array.)
As x is not const, you are free to modify it. x++ modifies it to point to the second element in the array.
When you then attempt to free the dynamically-allocated array (free(x)) you are passing free() a value that malloc did not give you. Hence the debugger triggering a break.
Either restore x to its prior value before attempting to free it, or use a different pointer variable to play arithmetic and leave x alone.