- Recall that:
- Operators operates on one or more operands (values) and return another value (kind of like a function)
- We will look at 2 very unusual operators in the C/C++ programming language that is not usually found in most other high level programming languages...
- The reference operator &:
- The reference Operator & operates on a variable and returns the address of that variable
Example:
- Suppose the variable a
is stored (starting) at
memory address 5000:
Then:
&a will evaulates to 5000
- C program that illustrates
the & operator:
/* ----------------------------------- Here are some (global) variables ----------------------------------- */ int i1, i2; short s1, s2; char c1, c2; float f1, f2; double d1, d2; int main(int argc, char *argv[]) { printf("Address of i1 = %u\n", &i1 ); // %u = unsigned integer printf("Address of i2 = %u\n", &i2 ); printf("Address of s1 = %u\n", &s1 ); printf("Address of s2 = %u\n", &s2 ); printf("Address of c1 = %u\n", &c1 ); printf("Address of c2 = %u\n", &c2 ); printf("Address of f1 = %u\n", &f1 ); printf("Address of f2 = %u\n", &f2 ); printf("Address of d1 = %u\n", &d1 ); printf("Address of d2 = %u\n", &d2 ); }Result:
Address of i1 = 134744 Address of i2 = 134748 (Notice: i2 located 4 bytes after i1) Address of s1 = 134752 Address of s2 = 134754 (Notice: s2 located 2 bytes after s1) Address of c1 = 134712 Address of c2 = 134713 (Notice: c2 located 1 bytes after c1) Address of f1 = 134736 Address of f2 = 134740 (Notice: f2 located 4 bytes after f1) Address of d1 = 134720 Address of d2 = 134728 (Notice: d2 located 8 bytes after d1)
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link(s) and
save in a scratch directory
- To compile: gcc ref-op1.c
- To run: ./a.out
- The dereference operator *:
- The dereference operator *
operates on an
number (= memory address)
and
- It returns the value (data) stored at that number (= memory address)
- The dereference operator *
operates on an
number (= memory address)
and
- Wait, there is a
problem:
- How many bytes
should the operation return ???
And How do you decode the bit pattern ???
Example:
- In other words:
- What (data) type of
value is stored at the memory location ???
- And ... how do we decode the value ?
- What (data) type of
value is stored at the memory location ???
Solution:
- We must also tell the C compiler the type of data that is stored at an address in memory !!!
- How many bytes
should the operation return ???
- Syntax:
(dataType*) addressmeaning:
- Tell the C compiler that
a
dataType typed
value is stored
at memory address
address
In fact, "(dataType*) address" means:
- address is an address of a variable of the type dataType.
- Tell the C compiler that
a
dataType typed
value is stored
at memory address
address
- Example:
In other words:
(int*) 5000 means: 5000 is an address of an int typed variable (short*) 5000 means: 5000 is an address of an short typed variable (float*) 5000 means: 5000 is an address of an float typed variable
- Now we apply the
deference operator * on an
address:
* ( (int*) 5000 ) means: returns the content at address 5000 as an integer * ( (short*) 5000 ) means: returns the content at address 5000 as a short integer * ( (float*) 5000 ) means: returns the content at address 5000 as a float
- Example:
int i1 = 1234; short s1 = -54; float f1 = 3.14; int main(int argc, char *argv[]) { printf("&i1 (address of int var i1) = %u\n", (unsigned) &i1 ); printf("int value at 6295584 = %d\n", * ( (int*)6295584 ) ); printf("i1 = %d\n\n", i1); printf("&s1 (address of short var s1) = %u\n", (unsigned) &s1 ); printf("short value at 6295588 = %d\n", * ( (short*)6295588 ) ); printf("s1 = %d\n\n", s1); printf("&f1 (address of float var f1) = %u\n", (unsigned) &f1 ); printf("float value at 6295592 = %f\n", * ( (float*)6295592 ) ); printf("f1 = %f\n\n", f1); }Output:
&i1 (address of int var i1) = 6295584 int value at 6295584 = 1234 i1 = 1234 &s1 (address of short var s1) = 6295588 short value at 6295588 = -54 s1 = -54 &f1 (address of float var i1) = 6295592 float value at 6295592 = 3.140000 f1 = 3.140000
- Explanation:
- The expression
(int*)6295584 tells the compiler to
treat the integer
6295584 as an
address of an INTEGER variable
- Thus, the dereference operator
* operating
of an
address of an INTEGER variable
will return an
4 byte INTEGER value
-
Basically (because the compiler knows that an integer is stored
in 4 consecutive bytes in memory),
the computer will fetch
4 bytes
from memory and
use the 2-complement encoding scheme
(which we will learn later in this course -
see click here)
to find the value represented.
- The expression
(int*)6295584 tells the compiler to
treat the integer
6295584 as an
address of an INTEGER variable
- Example Program:
(Demo above code)
- Prog file: click here
How to run the program:
- Right click on link(s) and
save in a scratch directory
- To compile: gcc deref-op1.c
- To run: ./a.out
Note:
- Try using a different type on the same address location !!!