Lesson 11.4. Incrementing pointers

Explanation

When a pointer is incremented (ptr+1) its address is advanced by the number of bytes of the pointed type. Let's consider the (classical) size of the following types :

Type	Size
`short`	2 bytes
`long`	8 bytes

The following code increases the pointed address by 2 bytes:

short *ptr;
...
// Increases the pointed address by 2 bytes (sizeof(short))
ptr++;

The following code increases the pointed address by 8 bytes:

long *ptr;
...
// Increases the pointed address by 8 bytes (sizeof(long))
ptr++;

Example

The following code displays the address of 4 pointers (*char, *short, *int and *long) before and after incrementation.

printf ("Before, *char  = %p\n",   pChar);
printf ("After, *char  = %p\n\n", pChar+1);

printf ("Before, *short = %p\n",   pShort);
printf ("After, *short = %p\n\n", pShort+1);

printf ("Before, *int   = %p\n",   pInt);
printf ("After, *int   = %p\n\n", pInt+1);

printf ("Before, *long  = %p\n",   pLong);
printf ("After, *long  = %p\n\n", pLong+1);

We check that the addresses are increased by the size of the pointed type:

Before, *char  = 0x7ffd8f06531b
After, *char  = 0x7ffd8f06531c (+1 byte)

Before, *short = 0x7ffd8f06530e
After, *short = 0x7ffd8f065310 (+2 bytes)

Before, *int   = 0x7ffd8f0652fc
After, *int   = 0x7ffd8f065300 (+4 bytes)

Before, *long  = 0x7ffd8f0652e8
After, *long  = 0x7ffd8f0652f0 (+8 bytes)

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Array pointers

The name of an array used without braces is a pointer to the first cell of the array. Consider the following statement:

int t[10];

t is equivalent to &t and &t[0]. This is the reason why tab1=tab2 does not copy the array tab2 into the array tab1. This type of assessment should not be used.

For arrays, the concept remains the same: we increase by the size of the pointed type. For example:

// 2 cells array
int tab[2] = {2,4};

// First cell of array
printf ("*tab     = tab[0] = %d\n", *tab);
// Second cell of array
printf ("*(tab+1) = tab[1] = %d\n", *(tab+1));

*ptr     = tab[0] = 2
*(ptr+1) = tab[1] = 4

We can see that the pointer, increased by 1, addresses the second cell of the array.

Try online

However, there is a subtlety for multidimensional arrays: the pointer advances by one in the main dimension (typically a row), not by one cell.

Exercises

Exercise 1

Write a program that

declares an array of characters of 4 rows by 6 columns;
displays the address of the array;
displays the address of the array increased by 1.

By how many bytes has the pointer advanced? What to conclude ?

Before : 0x7ffddefd68f0
After  : 0x7ffddefd68f6

Do the exercise

Solution

Exercise 2

The following statements are provided:

char text[10] = "Hello", *ptr;

Display each letter of text separated by a tab. You must not use:

square brackets [ and ] ;
additional variables;
the string.h library.

Use only pointers. Here is the expected result:

H   e   l   l   o

Do the exercise

Solution

Quiz

By how many bytes does a pointer incremented by 1 increase?

by 1 byte

by 2 bytes

by 4 bytes

it depends on the pointer type

Check

What does the following code display?

short tab[]={1,2,3};
*(tab+2)=4;
printf ("tab[2] = %d", tab[2]);

tab[2] = 1

tab[2] = 2

tab[2] = 3

tab[2] = 4

Check

Regarding the following program:

short tab1[]={1,2,3}, tab2[]={1,2,3};

if (tab1==tab2)
  printf ("Same\n");
else 
  printf ("Not the same\n");

the code doesn't compile.

the program displays : Same

the program displays : Not the same

one should never write such a horrible thing

Check

Lesson 11.3 Summary Lesson 11.5

Lesson 11.4. Incrementing pointers

Explanation

Example

Array pointers

Exercises

Exercise 1

Exercise 2

Quiz

See also