Pointers in C

Imagine that we have an int called i. Its address could be represented by the symbol &i. If the pointer is to be stored as a variable, it should be stored like this.

int *pi = &i; which is the same as int *pi; pi=&i;

int * is the notation for a pointer to an int. & is the operator which returns the address of its argument. When it is used, as in &i we say it is referencing i.
The opposite operator, which gives the value at the end of the pointer is *. An example of use, known as de-referencing pi, would be
i = *pi;

Take care not to confuse the many uses of the * sign; Multiplication, pointer declaration and pointer de-referencing.

This is a very confusing subject, so let us illustrate it with an example. The following function fiddle takes two arguments, x is an int while y is a pointer to int. It changes both values.

fiddle(int x, int *y)
{ printf(" Starting fiddle: x = %d, y = %dn", x, *y);
x++;
(*y)++; //get the value (integer) from &y, increase by 1, and then save back.
printf("Finishing fiddle: x = %d, y = %dn", x, *y);
}
since y is a pointer, we must de-reference it before incrementing its value.

A very simple program to call this function might be as follows.

main()
{ int i = 0;
int j = 0;

printf(" Starting main : i = %d, j = %dn", i, j);
printf("Calling fiddle nown");.
fiddle(i, &j);
printf("Returned from fiddlen");
printf("Finishing main : i = %d, j = %dn", i, j);
}

Note here how a pointer to int is created using the & operator within the call fiddle(i, &j);.
The result of running the program will look like this.

Starting main : i = 0 ,j = 0
Calling fiddle now
Starting fiddle: x = 0, y = 0
Finishing fiddle: x = 1, y = 1
Returned from fiddle
Finishing main : i = 0, j = 1

After the return from fiddle the value of i is unchanged while j, which was passed as a pointer, has changed.

To summarize, if you wish to use arguments to modify the value of variables from a function, these arguments must be passed as pointers, and de-referenced within the function.

Where the value of an argument isn’t modified, the value can be passed without any worries about pointers.