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Using an Array Name as a Pointer

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Using an array name as a pointer

An array name is really a pointer to the first element of the array. For example, the following is legal.
int b[100];    // b is an array of 100 ints.  
int* p;        // p is a pointer to an int.  
p = b;         // Assigns the address of first element of b to p.  
p = &b[0];     // Exactly the same assignment as above.  

Array name is a const pointer

When you declare an array, the name is a pointer, which cannot be altered. In the previous example, you could never make this assignment.
   p = b;   // Legal -- p is not a constant.     
	  b = p;   // ILLEGAL because b is a constant, altho the correct type.

Pointer arithmetic

"Meaningful" arithmetic operations are allowed on pointers.
  • Add or subtract integers to/from a pointer. The result is a pointer.
  • Subtract two pointers to the same type. The result is an int.
  • Multiplying, adding two pointers, etc. don't make sense.

Pointer addition and element size

When you add an integer to a pointer, the integer is multiplied by the element size of the type that the pointer points to.
// Assume sizeof(int) is 4.  
int b[100];  // b is an array of 100 ints.  
int* p;      // p is a a pointer to an int.  
p = b;       // Assigns address of first element of b. Ie, &b[0]  
p = p + 1;   // Adds 4 to p (4 == 1 * sizeof(int)). Ie, &b[1]  

Equivalence of subscription and dereference

Because of the way C/C++ uses pointers and arrays, you can reference an array element either by subscription or * (the unary dereference operator).
int b[100];  // b is an array of 100 ints.  
int* p;      // p is a a pointer to an int.  
p = b;       // Assigns address of first element of b. Ie, &b[0]  
*p = 14;     // Same as b[0] = 14  
p = p + 1;   // Adds 4 to p (4 == 1 * sizeof(int)). Ie, &b[1]  
*p = 22;     // Same as b[1] = 22;  

Example - Two ways to add numbers in an array

The first uses subscripts, the second pointers. They are equivalent.
int a[100]; 
 . . .  
int sum = 0;  
for (int i=0; i<100; i++) 
{     sum += a[i];  }  
   
int a[100]; 
 . . .  
int sum = 0;  
for (int* p=a; p<a+100; p++)
 {     sum += *p;  }  

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