How much overhead is there in calling a function in C++?

Question

A lot of literature talks about using inline functions to \"avoid the overhead of a function call\". However I haven\'t seen quantifiable data. What is the actual overhead o

Answer 1

Each new function requires a new local stack to be created. But the overhead of this would only be noticeable if you are calling a function on every iteration of a loop over a very large number of iterations.

Answer 2

It's worth pointing out that an inlined function increases the size of the calling function and anything that increases the size of a function may have a negative affect on caching. If you're right at a boundary, "just one more wafer thin mint" of inlined code might have a dramatically negative effect on performance.

---

If you're reading literature that's warning about "the cost of a function call," I'd suggest it may be older material that doesn't reflect modern processors. Unless you're in the embedded world, the era in which C is a "portable assembly language" has essentially passed. A large amount of the ingenuity of the chip designers in the past decade (say) has gone into all sorts of low-level complexities that can differ radically from the way things worked "back in the day."

Answer 3

For very small functions inlining makes sense, because the (small) cost of the function call is significant relative to the (very small) cost of the function body. For most functions over a few lines it's not a big win.

Answer 4

There is not much overhead at all, especially with small (inline-able) functions or even classes.

The following example has three different tests that are each run many, many times and timed. The results are always equal to the order of a couple 1000ths of a unit of time.

#include <boost/timer/timer.hpp>
#include <iostream>
#include <cmath>

double sum;
double a = 42, b = 53;

//#define ITERATIONS 1000000 // 1 million - for testing
//#define ITERATIONS 10000000000 // 10 billion ~ 10s per run
//#define WORK_UNIT sum += a + b
/* output
8.609619s wall, 8.611255s user + 0.000000s system = 8.611255s CPU(100.0%)
8.604478s wall, 8.611255s user + 0.000000s system = 8.611255s CPU(100.1%)
8.610679s wall, 8.595655s user + 0.000000s system = 8.595655s CPU(99.8%)
9.5e+011 9.5e+011 9.5e+011
*/

#define ITERATIONS 100000000 // 100 million ~ 10s per run
#define WORK_UNIT sum += std::sqrt(a*a + b*b + sum) + std::sin(sum) + std::cos(sum)
/* output
8.485689s wall, 8.486454s user + 0.000000s system = 8.486454s CPU (100.0%)
8.494153s wall, 8.486454s user + 0.000000s system = 8.486454s CPU (99.9%)
8.467291s wall, 8.470854s user + 0.000000s system = 8.470854s CPU (100.0%)
2.50001e+015 2.50001e+015 2.50001e+015
*/


// ------------------------------
double simple()
{
   sum = 0;
   boost::timer::auto_cpu_timer t;
   for (unsigned long long i = 0; i < ITERATIONS; i++)
   {
      WORK_UNIT;
   }
   return sum;
}

// ------------------------------
void call6()
{
   WORK_UNIT;
}
void call5(){ call6(); }
void call4(){ call5(); }
void call3(){ call4(); }
void call2(){ call3(); }
void call1(){ call2(); }

double calls()
{
   sum = 0;
   boost::timer::auto_cpu_timer t;

   for (unsigned long long i = 0; i < ITERATIONS; i++)
   {
      call1();
   }
   return sum;
}

// ------------------------------
class Obj3{
public:
   void runIt(){
      WORK_UNIT;
   }
};

class Obj2{
public:
   Obj2(){it = new Obj3();}
   ~Obj2(){delete it;}
   void runIt(){it->runIt();}
   Obj3* it;
};

class Obj1{
public:
   void runIt(){it.runIt();}
   Obj2 it;
};

double objects()
{
   sum = 0;
   Obj1 obj;

   boost::timer::auto_cpu_timer t;
   for (unsigned long long i = 0; i < ITERATIONS; i++)
   {
      obj.runIt();
   }
   return sum;
}
// ------------------------------


int main(int argc, char** argv)
{
   double ssum = 0;
   double csum = 0;
   double osum = 0;

   ssum = simple();
   csum = calls();
   osum = objects();

   std::cout << ssum << " " << csum << " " << osum << std::endl;
}

The output for running 10,000,000 iterations (of each type: simple, six function calls, three object calls) was with this semi-convoluted work payload:

sum += std::sqrt(a*a + b*b + sum) + std::sin(sum) + std::cos(sum)

as follows:

8.485689s wall, 8.486454s user + 0.000000s system = 8.486454s CPU (100.0%)
8.494153s wall, 8.486454s user + 0.000000s system = 8.486454s CPU (99.9%)
8.467291s wall, 8.470854s user + 0.000000s system = 8.470854s CPU (100.0%)
2.50001e+015 2.50001e+015 2.50001e+015

Using a simple work payload of

sum += a + b

Gives the same results except a couple orders of magnitude faster for each case.