Is there a performance difference between i++ and ++i in C++?

We have the question is there a performance difference between i++ and ++i in C?

What's the answer for C++?

Solution 1:

[Executive Summary: Use ++i if you don't have a specific reason to use i++.]

For C++, the answer is a bit more complicated.

If i is a simple type (not an instance of a C++ class), then the answer given for C ("No there is no performance difference") holds, since the compiler is generating the code.

However, if i is an instance of a C++ class, then i++ and ++i are making calls to one of the operator++ functions. Here's a standard pair of these functions:

Foo& Foo::operator++()   // called for ++i
{
    this->data += 1;
    return *this;
}

Foo Foo::operator++(int ignored_dummy_value)   // called for i++
{
    Foo tmp(*this);   // variable "tmp" cannot be optimized away by the compiler
    ++(*this);
    return tmp;
}

Since the compiler isn't generating code, but just calling an operator++ function, there is no way to optimize away the tmp variable and its associated copy constructor. If the copy constructor is expensive, then this can have a significant performance impact.

Solution 2:

Yes. There is.

The ++ operator may or may not be defined as a function. For primitive types (int, double, ...) the operators are built in, so the compiler will probably be able to optimize your code. But in the case of an object that defines the ++ operator things are different.

The operator++(int) function must create a copy. That is because postfix ++ is expected to return a different value than what it holds: it must hold its value in a temp variable, increment its value and return the temp. In the case of operator++(), prefix ++, there is no need to create a copy: the object can increment itself and then simply return itself.

Here is an illustration of the point:

struct C
{
    C& operator++();      // prefix
    C  operator++(int);   // postfix

private:

    int i_;
};

C& C::operator++()
{
    ++i_;
    return *this;   // self, no copy created
}

C C::operator++(int ignored_dummy_value)
{
    C t(*this);
    ++(*this);
    return t;   // return a copy
}

Every time you call operator++(int) you must create a copy, and the compiler can't do anything about it. When given the choice, use operator++(); this way you don't save a copy. It might be significant in the case of many increments (large loop?) and/or large objects.

Solution 3:

Here's a benchmark for the case when increment operators are in different translation units. Compiler with g++ 4.5.

Ignore the style issues for now

// a.cc
#include <ctime>
#include <array>
class Something {
public:
    Something& operator++();
    Something operator++(int);
private:
    std::array<int,PACKET_SIZE> data;
};

int main () {
    Something s;

    for (int i=0; i<1024*1024*30; ++i) ++s; // warm up
    std::clock_t a = clock();
    for (int i=0; i<1024*1024*30; ++i) ++s;
    a = clock() - a;

    for (int i=0; i<1024*1024*30; ++i) s++; // warm up
    std::clock_t b = clock();
    for (int i=0; i<1024*1024*30; ++i) s++;
    b = clock() - b;

    std::cout << "a=" << (a/double(CLOCKS_PER_SEC))
              << ", b=" << (b/double(CLOCKS_PER_SEC)) << '\n';
    return 0;
}

O(n) increment

Test

// b.cc
#include <array>
class Something {
public:
    Something& operator++();
    Something operator++(int);
private:
    std::array<int,PACKET_SIZE> data;
};


Something& Something::operator++()
{
    for (auto it=data.begin(), end=data.end(); it!=end; ++it)
        ++*it;
    return *this;
}

Something Something::operator++(int)
{
    Something ret = *this;
    ++*this;
    return ret;
}

Results

Results (timings are in seconds) with g++ 4.5 on a virtual machine:

Flags (--std=c++0x)       ++i   i++
-DPACKET_SIZE=50 -O1      1.70  2.39
-DPACKET_SIZE=50 -O3      0.59  1.00
-DPACKET_SIZE=500 -O1    10.51 13.28
-DPACKET_SIZE=500 -O3     4.28  6.82

O(1) increment

Test

Let us now take the following file:

// c.cc
#include <array>
class Something {
public:
    Something& operator++();
    Something operator++(int);
private:
    std::array<int,PACKET_SIZE> data;
};


Something& Something::operator++()
{
    return *this;
}

Something Something::operator++(int)
{
    Something ret = *this;
    ++*this;
    return ret;
}

It does nothing in the incrementation. This simulates the case when incrementation has constant complexity.

Results

Results now vary extremely:

Flags (--std=c++0x)       ++i   i++
-DPACKET_SIZE=50 -O1      0.05   0.74
-DPACKET_SIZE=50 -O3      0.08   0.97
-DPACKET_SIZE=500 -O1     0.05   2.79
-DPACKET_SIZE=500 -O3     0.08   2.18
-DPACKET_SIZE=5000 -O3    0.07  21.90

Conclusion

Performance-wise

If you do not need the previous value, make it a habit to use pre-increment. Be consistent even with builtin types, you'll get used to it and do not run risk of suffering unecessary performance loss if you ever replace a builtin type with a custom type.

Semantic-wise

• i++ says increment i, I am interested in the previous value, though.
• ++i says increment i, I am interested in the current value or increment i, no interest in the previous value. Again, you'll get used to it, even if you are not right now.

Knuth.

Premature optimization is the root of all evil. As is premature pessimization.

Solution 4:

It's not entirely correct to say that the compiler can't optimize away the temporary variable copy in the postfix case. A quick test with VC shows that it, at least, can do that in certain cases.

In the following example, the code generated is identical for prefix and postfix, for instance:

#include <stdio.h>

class Foo
{
public:

    Foo() { myData=0; }
    Foo(const Foo &rhs) { myData=rhs.myData; }

    const Foo& operator++()
    {
        this->myData++;
        return *this;
    }

    const Foo operator++(int)
    {
        Foo tmp(*this);
        this->myData++;
        return tmp;
    }

    int GetData() { return myData; }

private:

    int myData;
};

int main(int argc, char* argv[])
{
    Foo testFoo;

    int count;
    printf("Enter loop count: ");
    scanf("%d", &count);

    for(int i=0; i<count; i++)
    {
        testFoo++;
    }

    printf("Value: %d\n", testFoo.GetData());
}

Whether you do ++testFoo or testFoo++, you'll still get the same resulting code. In fact, without reading the count in from the user, the optimizer got the whole thing down to a constant. So this:

for(int i=0; i<10; i++)
{
    testFoo++;
}

printf("Value: %d\n", testFoo.GetData());

Resulted in the following:

00401000  push        0Ah  
00401002  push        offset string "Value: %d\n" (402104h) 
00401007  call        dword ptr [__imp__printf (4020A0h)] 

So while it's certainly the case that the postfix version could be slower, it may well be that the optimizer will be good enough to get rid of the temporary copy if you're not using it.