CS101: Instructions and Memory
A peek inside the machine

In order to understand how Java programs are executed, it's useful to have a mental model of the machine. For now, let's just think about the CPU (central processing unit) and the RAM (random access memory). This is a simplified abstract view:

The CPU has

registers that hold values
ALU (arithmetic and logic unit) to do simple computations (+, <, etc.)
control unit that knows how to set up the hardware to execute instructions:
- LOAD -- brings a particular memory location's value into a particular register
- STORE -- copies a value from a register into memory
- JUMP -- skips over some instructions (to implement conditional statements)
- ADD -- adds 2 registers together and puts result in another register
- etc.

Compilation and Execution

After you have written a Java program, the Java compiler translates your program into a series of instructions.
When you execute the program, the instructions are placed into the memory.
During execution:
Instruction fetch/decode
Instruction execution
Instruction fetch/decode
Instruction execution
etc.

The Execution Stack and Activation Records

The part of the computer's memory used by your program is divided into three parts:

program memory -- where the instructions of your program are stored
stack memory -- where parameters, return values, and variables inside procedures are stored
heap memory -- (we'll see what this is later)

Let's consider a small program execution and see what happens in the memory. Consider the following procedures:

int healthRisk(int cigarettesPerDay, int poundsOverWeight,
               int exerciseHoursPerWeek) {
  int smokingRisk = 20 * Math.sqrt(cigarettesPerDay);
  int weightRisk = poundsOverweight - 8;
  return smokingRisk + weightRisk - square(exerciseHoursPerWeek);
}

int square(i) {
  return i * i;
}

Let's look at the execution stack for the call healthRisk(4, 15, 3).

Call by Value

Let's think about what's going on in the example we just did. When a procedure is called, space is created on the stack for the

return value
parameter values
local variables declared in the procedure

Notice that the values of the parameters are copied into the activation records. This is call-by-value.

What happens to the calling procedure if we assign to a parameter within the called procedure? For example,

int foo(int x, int y) {
  int z = bar(x);
  return x + y + z;
}

int bar(int x) {
  x = 5 * x;
  if (x > 10)
    return x;
  else
    return -x;
}

What is the return value from foo(3, 10)?

Notice that the parameter x to foo is different from the parameter x to bar.
Local names are used within a procedure.
When we modify x inside bar, the x in foo is unaffected.
Also, notice that even upon return, the value of x in foo is unchanged -- the parameter value is not copied back on return.
Also, upon return, the local variables and parameters of the procedure "go away."

To summarize, this is the semantics of call-by-value: The value of the actual parameter is copied into the formal parameter and used locally. Call-by-value is the parameter passing mechanism used by Java. (Some languages, such as C++, permit call-by-reference, where the called procedure uses a variable provided by the calling procedure. With call-by-reference, changes to the value of the variable are visible to the calling procedure upon return.)

Let's do one more example:

int foo(int x, int y, int z) {
  return z + bar(x, y);
}

int bar(int z, int w) {
  int x = z + w;
  z = square(z);
  return x + z;
}

int square(int x) {
  return x * x;
}

What is the return value of foo(3, 4, 5)?

Kennneth J. Goldman

Last modified: Mon Jan 20 15:50:36 CST

CS101: Instructions and Memory A peek inside the machine

Compilation and Execution

The Execution Stack and Activation Records

Call by Value

CS101: Instructions and Memory
A peek inside the machine