Pointer arithmetic

Review of reference (pointer) variables

Recall that:

int i; // An integer variable int a[10]; // An integer array int *p; // A reference variable to an int variable // p can contain an address of an int variable

Review: access variables through a reference variable

Consider these 2 kinds of int variables:
Illustrated:
Notice that:

Therefore, we can use an int *p variable to access any one of these int typed variables:

int *p; p = &i; *p will access i p = &a[0]; *p will access a[0] p = &a[1]; *p will access a[1]

Pointer arithmentic: the + operator on reference variables

Important fact about pointer/reference variables:

A reference/pointer variable always has an associated data type:

Example:

int *p; // p must point to an int typed variable float *q; // q must point to an float typed variable

Pointer addition: what does it mean to add an integer to a pointer value

The expression:

referenceVariable + integer

evaluates (returns) to the address

referenceVariable + integer*sizeof( (*referenceVariable) )

Comment:

If you remember the lecture on how to access array elements in assembler code ( click here), then you should already understand what adding an integer to a pointer value achieves.

Example:

Initially, the variable p points to the variable a[0]:

int a[10]; int *p; p = &a[0];

Then the expression p + 1 will evaluates to:

p + 1 ===> &a[0] + 1*sizeof( (*p) ) // (*p) is an int ! <===> &a[0] + 1*4 <===> &a[0] + 4

The expression p + 2 will evaluates to:

p + 2 ===> &a[0] + 1*sizeof( (*p) ) // (*p) is an int ! <===> &a[0] + 2*4 <===> &a[0] + 8

Example C progra:

int main(int argc, char *argv[]) { int a[10]; int *p; p = &a[0]; // p points to variable a[0] printf("p = %u, &a[%d] = %u\n", p, 0, &a[0] ); printf("p + 1 = %u, &a[%d] = %u\n", p+1, 1, &a[1] ); printf("p + 2 = %u, &a[%d] = %u\n", p+2, 2, &a[2] ); printf("p + 3 = %u, &a[%d] = %u\n", p+3, 3, &a[3] ); }

Output of C program:

p = 4290769444, &a[0] = 4290769444 (increased by 4 !!!) p + 1 = 4290769448, &a[1] = 4290769448 p + 2 = 4290769452, &a[2] = 4290769452 p + 3 = 4290769456, &a[3] = 4290769456

Example Program: (Demo above code)
- Prog file: click here
How to run the program:

The following figure shows the result of the p+1 and p+2 more clearly:

In other words:

If p points to a[0] then: p + 1 = &a[0] + 4 = the address of the array element a[1] !! p + 2 = &a[0] + 8 = the address of the array element a[2] !!

In general:

The expression:
evaluates to:

The − operator on reference variables

The − operator on reference variables has a similar effect as the + operator:

Suppose, the variable p points to the variable a[5]:

int a[10]; int *p; p = &a[5];

Then the expression p - 1 will evaluates to:

p - 1 ===> &a[5] - 1*sizeof( (*p) ) // (*p) is an int ! <===> &a[5] - 1*4 <===> &a[5] - 4

The expression p - 2 will evaluates to:

p - 2 ===> &a[5] - 1*sizeof( (*p) ) // (*p) is an int ! <===> &a[5] - 2*4 <===> &a[5] - 8

In other words:

If p points to a[5] then: p - 1 = &a[5] - 4 = the address of the array element a[4] !! p - 2 = &a[5] - 8 = the address of the array element a[3] !!

In general:

The expression:
evaluates to: