c++ Why is memory allocated when passing a string literal to a function?

Question

I have his code:

int setAttrib(const string& name, int components){    
    // here I don't even touch 'name'
    if(components == 2) return 3;
    else return 1;
}

And I call the function this way:

setAttrib("position", 3);

I'm profiling memory with xcode profiler and in the function call std::string is making an allocation.
Why is that?

EDIT:

What's the best way to avoid that allocation? since I'm calling that function a lot, and in about 10 seconds of time I end up allocating about 10MB in that line.

Thanks.

It depends on your compiler. Some compilers will not allocate dynamic memory for such a short string. [http://stackoverflow.com/questions/1466073/how-is-stdstring-implemented](http://stackoverflow.com/questions/1466073/how-is-stdstring-implemented) — Bo Persson, May 30 '11 at 11:38

score 14 · Accepted Answer · answered May 30 '11 at 10:56

14

You ask for a const string&, but pass in a const char*. The compiler thus needs to create a temporary object of the correct type.

The fact that "position" is not an std::string but a char const* is more of a historical accident (inherited from C, when there was no string class in C++) than a design decision, but something to keep in mind nevertheless.

answered May 30 '11 at 10:56

Christopher Creutzig

8,656
35
45

1

+1 Since this is the only response which seems to hit on the key issue: a string literal doesn't have type `std::string`. – James Kanze May 30 '11 at 10:57
workaround for "multiple" allocations, is to make static std::string attrName("position") one line before setAttrib("posisition"...) and replace const literal with attrName variable – Kamil Klimek May 30 '11 at 11:40
@Kamil: you could even define it in the namespace, as a constant. – Matthieu M. May 30 '11 at 12:02

Alexander Gessler · Answer 2 · 2011-05-30T11:00:22.380

Because std::string typically allocates heap memory to hold the string. In this case, a std::string is implicitly constructed from a string literal (which by itself resides in static storage). Some string implementations use a small buffer to serve small strings, but this doesn't seem to be the case here and is implementation-dependent anyway.

It doesn't matter that name is not used - basically, setAttrib("position", 3) is shorthand for setAttrib(std::string("position"), 3);, so when control enters setAttrib, the memory has already been allocated (of course, in your isolated code sample it would be possible for a compiler to inline getAttrib and then drop the string construction at all, but this is a compiler optimization, not a language feature).

Note that temporary objects created during function invocation are automatically destructed when the function returns, so there is no memory leaked.

"and then drop the string construction at all" constructors are never optimized out. They could have side effects a la RAII. — IronMensan, May 30 '11 at 14:20

score 5 · Answer 3 · answered May 30 '11 at 10:55

In order to call the function the compiler needs to construct all parameters, const string& name included, the only way to do that in your case (you pass a string literal instead) is to construct a temporary std::string and this requires a heap memory allocation in most implementations. It doesn't matter if you use the value inside the function or not.

c++ Why is memory allocated when passing a string literal to a function?

3 Answers3