C++ string construct performance

Question

I was wondering if below code has any difference between #1, #2 and #3 when passing a string to a function (eg. insert into a vector). Especially when dealing with millions of strings in the code.

std::vector<std::string> v;
std::string s("foo");
int i = 1;
v.push_back( s + "bar" + boost::lexical_cast<std::string>(i) );  // #1
v.push_back( std::string(s + "bar" + boost::lexical_cast<std::string>(i)) );  // #2
std::string s2 = s + "bar" + boost::lexical_cast<std::string>(i);
v.push_back(s2);  // #3

orlp · Accepted Answer · 2012-09-25 21:53:17Z

1

With a decent optimizer there should be no difference between #1 and #2, but #1 gives you the biggest chance that the compiler actually does temporary elimination.

However, #3 contains a named temporary, so it probably needs a great optimizer to detect that it's just that - a temporary.

answered Sep 25, 2012 at 21:53

orlp

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3 Comments

Edward Strange Over a year ago

push_back takes T as const&. The compiler will be binding the temporary created by the expression or the one passed in as a variable. It won't make a bit of difference.

ildjarn Over a year ago

@CrazyEddie : In C++11 there's a big difference – #1 and #2 are rvalues and will be moved into the vector automatically, #3 would need to be moved explicitly.

Stan Over a year ago

What if I need to loop through these code million times, is there an efficient way to reuse the memory space for the temporary string? Just like the way C# does with StringBuilder class. It only creates once, and reuses the memory over and over again.

Edward Strange · Accepted Answer · 2012-09-25 21:54:17Z

1

Probably not. They should be equivalent.

One way you could get performance gain though is by pre-allocating your string:

std::string s;
s.reserve(10); // or so...
s.append("foobar").append(lexical_cast<std::string>(i));

You probably would not gain enough performance here to make it worth the hassle though.

answered Sep 25, 2012 at 21:54

Edward Strange

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