How is default argument to a method implemented in C++?

C++ allows a function to assign an argument a default value when no argument is specified in a call to that function. The third argument of following function will have “0” when it is not passed by the caller.

int sum(int num1, int num2, int num3 = 0)

{

       return num1 + num2 + num3;

}

When a method is called, all the arguments are pushed on stack and method pops them from stack and copy values in the formal arguments.

Let’s see the dis-assembly code generated for a method call with all the arguments:

sum (1,2,3);

       // Here is the dis-assembly code for passing arguments to sum method

push        3 //push value 3 on stack  

push        2 //push value 2 on stack      

push        1 //push value 1 on stack      

call        sum (41123Fh) // call sum method

Here, we can see that all the three values are pushed on the stack.

Now see the dis-assembly code generated for a method call when no value to supplied for last argument:

sum(1,2);

       // Here is the dis-assembly code for passing arguments to sum method

push        0 //Here 0 as default value for last argument

push        2 //push value 2 on stack      

push        1 //push value 1 on stack      

call        sum (41123Fh) // call sum method

We can see that “push 0” assembly  code is pushing “0” (specified in method’s definition) as default value for last argument “num3”.