Q:

Accessing array by pointers

let int * ptr,array[10]; ptr= array;.  Now each of the memory cells in array's contiguous locations have a fixed size.  If the address of first cell is 1234 then the next cell must be at 1238 address.  But we access it using a pointer as *(ptr+1). 
I am confused regarding this.  Any sources or answers?  Thanks. 

A:

From the C11 standard: §6.5.2.1

A postﬁx expression followed by an expression in square brackets [] is a subscripted
  designation of an element of an array object.  The deﬁnition of the subscript operator []
  is that E1[E2] is identical to (*((E1)+(E2))).  Because of the conversion rules that
  apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the
  initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th
  element of E1 (counting from zero). 

E.G. what you're doing is basically what [] already does
Also (same standard) explains why pointers increment as you noticed: §6.5.6

When an expression that has integer type is added to or subtracted from a pointer, the
  result has the type of the pointer operand.  If the pointer operand points to an element of
  an array object, and the array is large enough, the result points to an element offset from
  the original element such that the difference of the subscripts of the resulting and original
  array elements equals the integer expression.  In other words, if the expression P points to
  the i-th element of an array object, the expressions (P)+N (equivalently, N+(P)) and
  (P)-N (where N has the value n) point to, respectively, the i+n-th and i−n-th elements of
  the array object, provided they exist.  Moreover, if the expression P points to the last
  element of an array object, the expression (P)+1 points one past the last element of the
  array object, and if the expression Q points one past the last element of an array object,
  the expression (Q)-1 points to the last element of the array object.  If both the pointer
  operand and the result point to elements of the same array object, or one past the last
  element of the array object, the evaluation shall not produce an overﬂow; otherwise, the
  behavior is undeﬁned.  If the result points one past the last element of the array object, it
  shall not be used as the operand of a unary * operator that is evaluated. 

A:

The compiler knows that ptr is a pointer to a 4-byte type, so it knows that ptr+1 is 4 bytes further on in the memory. 
If you think for a moment, it's clear that this must be so, as otherwise you couldn't write portable code without knowing the size of (for instance) an integer on your system. 
Further, array indexing is exactly pointer arithmetic under the covers - that is, array[3] is exactly the same as *(array + 3)

