CODEWITHSUNDEEP
parameter-passing
name_masked
name_masked

Reputation: 9793

Passing expression as a parameter in Call by reference

All,

When we are passing an expression as a parameter, how does the evaluation occur? Here is a small example. This is just a pseudocode kind of example:

f (x,y)
{
    y = y+1;
    x = x+y;
}
main()
{
    a = 2; b = 2;
    f(a+b, a)
    print a;
}

When accessing variable x in f, does it access the address of the temp variable which contains the result of a+b or will it access the individual addresses of a and b and then evaluate the value of a+b

Please help.

Regards, darkie15

Upvotes: 0

Views: 1758

Answers (2)

sqykly
sqykly

Reputation: 21

In C or C++, as long as x and y are not pointers (in which case the expression is not useful anyway), they are both evaluated before the function call and the VALUE of the result is pushed on the stack. There are no references involved, at all.

All parameters in C and C++ are always passed by value. If a reference type (eg int*, int&) is passed to the function, the VALUE of the reference is passed. While the referenced object may be changed by accessing eg *x within the function, the value of the reference still cannot be changed, because C and C++ parameters are always always always passed by value only.

EDIT: an exception in C and C++ is the case in which some overloaded operator is defined like the following:

T* operator+ (L lhs, R rhs) {return new T(lhs, rhs);}

and x is an L, and y is an R. In this case, the value of the T* generated by the function is pushed on the stack as a parameter. Don't write code like that, it confuses other programmers =D.

Upvotes: 1

anon
anon

Reputation:

Somewhat language dependent, but in C++

f(a+b, a)

evaluates a + b and and pushes the result of evaluation onto the stack and then passes references to this value to f(). This will only work if the first parameter is of f() is s const reference, as temporary objects like the result of a + b can only be bound to const references.

Upvotes: 1

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