Efficiency of Postincrement V.S. Preincrement in C++

Efficiency of postincrement v.s. preincrement in C++

pre-increment introduces a data dependency into your code -- the CPU must wait for the increment operation to be completed before its value can be used in the expression."

Is this true?

It is mostly true - although perhaps overly strict. Pre increment doesn't necessarily introduce a data dependency - but it can.

A trivial example for exposition:

a = b++ * 2;

Here, the increment can be executed in parallel with the multiplication. The operands of both the increment and the multiplication are immediately available and do not depend on the result of either operation.

Another example:

a = ++b * 2;

Here, the multiplication must be executed after the increment, because one of the operands of the multiplication depends on the result of the increment.

Of course, these statements do slightly different things, so the compiler might not always be able to transform the program from one form to the other while keeping the semantics the same - which is why using the post-increment might make a slight difference in performance.

A practical example, using a loop:

for(int i= 0; arr[i++];)
    count++;

for(int i=-1; arr[++i];) // more typically: (int i=0; arr[i]; ++i;)
    count++;

One might think that the latter is necessarily faster if they reason that "post-increment makes a copy" - which would have been very true in the case of non-fundamental types. However, due to the data dependency (and because int is a fundamental type with no overload function for increment operators), the former can theoretically be more efficient. Whether it actually is depends on the CPU architecture, and the ability of the optimizer.

For what it's worth - in a trivial program, on x86 arch, using g++ compiler with optimization enabled, the above loops had identical assembly output, so they are perfectly equivalent in that case.

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Rules of thumb:

If the counter is a fundamental type and the result of increment is not used, then it makes no difference whether you use post/pre-increment.

If the counter is not a fundamental type and the result of the increment is not used and optimizations are disabled, then pre-increment may be more efficient. With optimizations enabled, there is usually no difference.

If the counter is a fundamental type and the result of increment is used, then post-increment can theoretically be marginally more efficient - in some CPU architecture - in some context - using some compiler.

If the counter is not a fundamental type and the result of the increment is used, then pre-increment is typically faster than post-increment. Also, see R Sahu's answer regarding this case.

Preincrement faster than postincrement in C++ - true? If yes, why is it?

Post-increment usually involves keeping a copy of the previous value around and adds a little extra code. Pre-increment simply does it's job and gets out of the way. I typically pre-increment unless the semantics would change and post-increment is actually necessary.

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

Executive summary: No.

i++ could potentially be slower than ++i, since the old value of i
might need to be saved for later use, but in practice all modern
compilers will optimize this away.

We can demonstrate this by looking at the code for this function,
both with ++i and i++.

$ cat i++.c
extern void g(int i);
void f()
{
    int i;

    for (i = 0; i < 100; i++)
        g(i);

}

The files are the same, except for ++i and i++:

$ diff i++.c ++i.c
6c6
<     for (i = 0; i < 100; i++)
---
>     for (i = 0; i < 100; ++i)

We'll compile them, and also get the generated assembler:

$ gcc -c i++.c ++i.c
$ gcc -S i++.c ++i.c

And we can see that both the generated object and assembler files are the same.

$ md5 i++.s ++i.s
MD5 (i++.s) = 90f620dda862cd0205cd5db1f2c8c06e
MD5 (++i.s) = 90f620dda862cd0205cd5db1f2c8c06e

$ md5 *.o
MD5 (++i.o) = dd3ef1408d3a9e4287facccec53f7d22
MD5 (i++.o) = dd3ef1408d3a9e4287facccec53f7d22

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

[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.

Does Pre and post increment/decrement operators in C++ have same performance in a loop?

As noted in this answer, pre is faster. They are only the same when you are dealing with primitives (int, char, etc.). In this case, they are calling overloaded operators.

Which one is faster post increment or pre-increment?

There are few answers out there:

http://www.geekinterview.com/question_details/14424

http://www.digitalpeer.com/id/where

but this question is a duplicate

Preincrement faster than postincrement in C++ - true? If yes, why is it?

---andrew

Difference between pre-increment and post-increment in a loop?

a++ is known as postfix.

add 1 to a, returns the old value.

++a is known as prefix.

add 1 to a, returns the new value.

C#:

string[] items = {"a","b","c","d"};
int i = 0;
foreach (string item in items)
{
    Console.WriteLine(++i);
}
Console.WriteLine("");

i = 0;
foreach (string item in items)
{
    Console.WriteLine(i++);
}

Output:

1
2
3
4

0
1
2
3

foreach and while loops depend on which increment type you use. With for loops like below it makes no difference as you're not using the return value of i:

for (int i = 0; i < 5; i++) { Console.Write(i);}
Console.WriteLine("");
for (int i = 0; i < 5; ++i) { Console.Write(i); }

0 1 2 3 4

0 1 2 3 4

If the value as evaluated is used then the type of increment becomes significant:

int n = 0;
for (int i = 0; n < 5; n = i++) { }

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