c - Is an int pointer an array of ints? -

I'm following a few examples which declare an integer pointer

  Int * myInt;   
 and then turn that pointer into an array  
  myInt = (int *) malloc (1024);   
 Checks it  
  myInt [0] = 5; Cout & lt; & Lt; Minute [0]; // print 5 myint [1] = 7; Cout & lt; & Lt; Mint [1]; // print 7   
 I thought an integer indicator was an indicator for an int and never at all. I know that the signal to wire only indicates the first letter of the string, but it appears that there is something similar in the same kind of things here. But then what we want is an array of things, why not create an array of inset instead of an indicator rather than an int?  
 The way I work in C, C ++ is in a C ++ file but the related code is in C.  
  
 
 
 There is an optional syntax for accessing items indicated by a pointer - square brackets This syntax gives you points of Pointer Allows access to data by means such that an indicator was an array (of course, the indicators are  not  arrays)  is a [i]  an expression, which is an alternative form of  * (a + i)   * ).  
 When you allocate the dynamic storage and assign it to  myInt , then you can use the pointer as a dynamic array that can resize at runtime:  
  myInt = malloc (1024 * sizeof (int)); // C does not need an artist, only for C ++ (int i = 0; i! = 1024; i ++) {myInt [i] = i; // square bracket syntax for use} (int i = 0; i! = 1024; i ++) {printf ("% d", * (myInt + i)); // Use the equivalent indicator syntax)   
  
  *  Incidentally, write the comma of  +  Returns  4 [array]  instead of  array [4] ; Do not do this!