accessing the (n+1)th the element of an array of size n by using pointer

Question

In an array A[n] of size n we can access the element A[n+1] by writting *(A+n+1) . Accessing the all the elements before and after the index can be done in 0(1) time complexity.So, why we will use a linked list if we can access any no of elements after the end point of an array . The   size of each node is also larger in case of linked list.

Answer 1

why we use link list is that there are various thing u need to know . so one by one . there is something called bound checking which is not a feature of c but a feature of c++ . whenever u are going to acess data out of array it is going to give u error. so why we need to go for bound checking because to maintain order of one acess time the array must be stored in sequencial order. and if we say a[5] system allocate all the memory in continious manner to the array . while the memory after that may not be free. if we just go writing it . it may overwrite the data of other program . whereas linklist are dynamic in nature . so no need of continous allocation so. array are not used .it may have worked. but this type of programming is a hazard to system.

Comments

when we access data out of array it shows an error if we are using the syntax or array a[i] but it does nt show an error if we access out of array elements using pointer.for example = *(a+i). is nt it true?? i m satisfied with the rest of the point ...serial access is a problem,memory overwriting can be a problem...