C++ – The cost of passing by shared_ptr

cperformanceshared-ptr

I use std::tr1::shared_ptr extensively throughout my application. This includes passing objects in as function arguments. Consider the following:

class Dataset {...}

void f( shared_ptr< Dataset const > pds ) {...}
void g( shared_ptr< Dataset const > pds ) {...}
...

While passing a dataset object around via shared_ptr guarantees its existence inside f and g, the functions may be called millions of times, which causes a lot of shared_ptr objects being created and destroyed. Here's a snippet of the flat gprof profile from a recent run:

Each sample counts as 0.01 seconds.
  %   cumulative   self              self     total
 time   seconds   seconds    calls   s/call   s/call  name
  9.74    295.39    35.12 2451177304     0.00     0.00  std::tr1::__shared_count::__shared_count(std::tr1::__shared_count const&)
  8.03    324.34    28.95 2451252116     0.00     0.00  std::tr1::__shared_count::~__shared_count()

So, ~17% of the runtime was spent on reference counting with shared_ptr objects. Is this normal?

A large portion of my application is single-threaded and I was thinking about re-writing some of the functions as

void f( const Dataset& ds ) {...}

and replacing the calls

shared_ptr< Dataset > pds( new Dataset(...) );
f( pds );

with

f( *pds );

in places where I know for sure the object will not get destroyed while the flow of the program is inside f(). But before I run off to change a bunch of function signatures / calls, I wanted to know what the typical performance hit of passing by shared_ptr was. Seems like shared_ptr should not be used for functions that get called very often.

Any input would be appreciated. Thanks for reading.

-Artem

Update: After changing a handful of functions to accept const Dataset&, the new profile looks like this:

Each sample counts as 0.01 seconds.
  %   cumulative   self              self     total
 time   seconds   seconds    calls   s/call   s/call  name
  0.15    241.62     0.37 24981902     0.00     0.00  std::tr1::__shared_count::~__shared_count()
  0.12    241.91     0.30 28342376     0.00     0.00  std::tr1::__shared_count::__shared_count(std::tr1::__shared_count const&)

I'm a little puzzled by the number of destructor calls being smaller than the number of copy constructor calls, but overall I'm very pleased with the decrease in the associated run-time. Thanks to all for their advice.

Best Answer

Always pass your shared_ptr by const reference:

void f(const shared_ptr<Dataset const>& pds) {...} 
void g(const shared_ptr<Dataset const>& pds) {...}

Edit: Regarding the safety issues mentioned by others:

When using shared_ptr heavily throughout an application, passing by value will take up a tremendous amount of time (I've seen it go 50+%).
Use const T& instead of const shared_ptr<T const>& when the argument shall not be null.
Using const shared_ptr<T const>& is safer than const T* when performance is an issue.

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C++ – the difference between #include and #include “filename”

In practice, the difference is in the location where the preprocessor searches for the included file.

For #include <filename> the preprocessor searches in an implementation dependent manner, normally in search directories pre-designated by the compiler/IDE. This method is normally used to include standard library header files.

For #include "filename" the preprocessor searches first in the same directory as the file containing the directive, and then follows the search path used for the #include <filename> form. This method is normally used to include programmer-defined header files.

A more complete description is available in the GCC documentation on search paths.

C++ – What are the differences between a pointer variable and a reference variable in C++

A pointer can be re-assigned:

int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);

A reference cannot be re-bound, and must be bound at initialization:

int x = 5;
int y = 6;
int &q; // error
int &r = x;

A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary & operator and a certain amount of space that can be measured with the sizeof operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
```
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;

assert(p == p2); // &x == &r
assert(&p != &p2);
```

You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.

int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;

**pp = 2;
pp = &q; // *pp is now q
**pp = 4;

assert(y == 4);
assert(x == 2);

A pointer can be assigned nullptr, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr, but this is undefined and will not behave consistently.

/* the code below is undefined; your compiler may optimise it
 * differently, emit warnings, or outright refuse to compile it */

int &r = *static_cast<int *>(nullptr);

// prints "null" under GCC 10
std::cout
    << (&r != nullptr
        ? "not null" : "null")
    << std::endl;

bool f(int &r) { return &r != nullptr; }

// prints "not null" under GCC 10
std::cout
    << (f(*static_cast<int *>(nullptr))
        ? "not null" : "null")
    << std::endl;

You can, however, have a reference to a pointer whose value is nullptr.

Pointers can iterate over an array; you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access its members whereas a reference uses a ..
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
```
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
```
This makes const & more convenient to use in argument lists and so forth.

Best Answer

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