Reputation: 1953
I am a novice programmer and have only briefly covered the anatomy of a function call (setting up the stack, etc.). I can write a function two different ways and I'm wondering which (if either) is more efficient. This is for a finite element program so this function could be called several thousand times. It is using the linear algebra library Aramdillo.
First way:
void Q4::stiffness(mat &stiff)
{
stiff.zeros; // sets all elements of the matrix to zero
// a bunch of linear algebra calculations
// ...
stiff *= h;
}
int main()
{
mat elementStiffness(Q4__DOF, Q4__DOF);
mat globalStiffness(totalDOF, totalDOF);
for (int i = 0; i < reallyHugeNumber; i++)
{
elements[i].stiffness(&elementStiffness, PSTRESS);
assemble(&globalStiffness, &elementStiffness);
}
return 0;
}
Second way:
mat Q4::stiffness()
{
mat stiff(Q4__DOF, Q4__DOF); // initializes element stiffness matrix
// a bunch of linear algebra calculations
// ...
return stiff *= h;
}
int main()
{
mat elementStiffness(Q4__DOF, Q4__DOF);
mat globalStiffness(totalDOF, totalDOF);
for (int i = 0; i < reallyHugeNumber; i++)
{
elementStiffness = elements[i].stiffness(PSTRESS);
assemble(&globalStiffness, &elementStiffness);
}
return 0;
}
I think what I'm asking is: using the second way is mat stiff
pushed to the stack and then copied into elementStiffness
? Because I imagine the matrix being pushed to the stack and then being copied is much more expensive than passing a matrix be reference and setting its elements to zero.
Upvotes: 3
Views: 144
Reputation: 924
Passing a variable by reference and doing your calculations on that variable is a lot cheaper. When c++ returns a variable, it pretty much copies it twice.
First inside the function, and then it calls the copy constructor or assignment operator, depending on if the value is being assigned to a new variable or to an existing variable, to initialize the variable. If you have a user-defined variable with a long list of internal state variables then this assignment operation is going to take a big chunk of the operator's processing time.
EDIT#1: I forgot about c++11 and the std::move. Many compilers can optimize functions like this so they can use std::move and instead of copying an lvaue it can copy an rvalue which is just the memory location.
Upvotes: 3
Reputation: 40
On the surface, I think the second way will be much more expensive as it both constructs a new mat and copies it to the stack on every call. Of course that depends a bit on how often the mat construction takes place in the first way.
That said, I think the best thing to do is setup an experiment and test to make sure (agreeing with the suggestion to research).
Upvotes: 0