Turning temporary stringstream to c_str() in single statement

Question

Consider the following function:

void f(const char* str);

Suppose I want to generate a string using stringstream and pass it to this function. If I want to do it in one statement, I might try:

f((std::ostringstream() << "Value: " << 5).str().c_str()); // error

This gives an error: 'str()' is not a member of 'basic_ostream'. OK, so operator<< is returning ostream instead of ostringstream - how about casting it back to an ostringstream?

1) Is this cast safe?

f(static_cast<std::ostringstream&>(std::ostringstream() << "Value: " << 5).str().c_str()); // incorrect output

Now with this, it turns out for the operator<<("Value: ") call, it's actually calling ostream's operator<<(void*) and printing a hex address. This is wrong, I want the text.

2) Why does operator<< on the temporary std::ostringstream() call the ostream operator? Surely the temporary has a type of 'ostringstream' not 'ostream'?

I can cast the temporary to force the correct operator call too!

f(static_cast<std::ostringstream&>(static_cast<std::ostringstream&>(std::ostringstream()) << "Value: " << 5).str().c_str());

This appears to work and passes "Value: 5" to f().

3) Am I relying on undefined behavior now? The casts look unusual.

---

I'm aware the best alternative is something like this:

std::ostringstream ss;
ss << "Value: " << 5;
f(ss.str().c_str());

...but I'm interested in the behavior of doing it in one line. Suppose someone wanted to make a (dubious) macro:

#define make_temporary_cstr(x) (static_cast<std::ostringstream&>(static_cast<std::ostringstream&>(std::ostringstream()) << x).str().c_str())

// ...

f(make_temporary_cstr("Value: " << 5));

Would this function as expected?

Answer 1

A temporary cannot be passed as a non-const reference to a function, that's why it does not find the correct streaming operator and instead takes the one with void* argument (it is a member function and thus calling it on a temporary is OK).

What comes to getting around the limitations by casting, I have a feeling that it is in fact UB, but I cannot say for sure. Someone else will surely quote the standard.

Answer 2

You cannot cast the temporary stream to std::ostringstream&. It is ill-formed (the compiler must tell you that it is wrong). The following can do it, though:

f(static_cast<std::ostringstream&>(
  std::ostringstream().seekp(0) << "Value: " << 5).str().c_str());

That of course is ugly. But it shows how it can work. seekp is a member function returning a std::ostream&. Would probably better to write this generally

template<typename T>
struct lval { T t; T &getlval() { return t; } };

f(static_cast<std::ostringstream&>(
  lval<std::ostringstream>().getlval() << "Value: " << 5).str().c_str());

The reason that without anything it takes the void*, is because that operator<< is a member-function. The operator<< that takes a char const* is not.

Answer 3

If you like one_lined statements you can write:

// void f(const char* str); 
f(static_cast<ostringstream*>(&(ostringstream() << "Value: " << 5))->str());

However you should prefer easier to maintain code as:

template <typename V>
  string NumberValue(V val)
  {
     ostringstream ss;
     ss << "Value: " << val;
     return ss.str();
  }
f(NumberValue(5));

cheers, AR