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?
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.
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.
This can be done using a C++11 lambda function.
#include <iostream>
#include <sstream>
void f(const char * str)
{
std::cout << str << std::endl;
}
std::string str(void (*populate)(std::ostream &))
{
std::ostringstream stream;
populate(stream);
return stream.str();
}
int main(int argc, char * * args)
{
f(str([](std::ostream & ss){ ss << "Value: " << 5; }).c_str());
return 0;
}
// g++ -std=c++11 main.cpp -o main
// ./main
// Value: 5
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));
I use something a bit like this for logging.
#include <sstream>
using namespace std;
const char *log_text(ostringstream &os, ostream &theSame)
{
static string ret; // Static so it persists after the call
ret = os.str();
os.str(""); // Truncate so I can re-use the stream
return ret.c_str();
}
int main(int argc, char **argv)
{
ostringstream ss;
cout << log_text(ss, ss << "My first message") << endl;
cout << log_text(ss, ss << "Another message") << endl;
}
Output:
My first message
Another message