问题
Today I made a small typo in my program, and was wandering why I wasn't getting any output, although the program compiled fine. Basically it reduces to this:
#include <iostream>
int main()
{
std::cout < "test"; // no << but <
}
I have absolutely no idea what kind of implicit conversion is performed here so the program still compiles (both g++4.9.2 and even g++5). I just realized that clang++ rejects the code. Is there a conversion to void*
being performed (cannot think of anything else)? I remember seeing something like this, but I thought it was addressed in g++5, but this doesn't seem to be the case.
EDIT: I was not compiling with -std=c++11
, so the code was valid in pre-C++11 (due to conversion to void*
of ostream
). When compiling with -std=c++11
g++5 rejects the code, g++4.9 still accepts it.
回答1:
Yes, the compiler is converting cout
to a void*
. If you use the -S
switch to get the code's disassembly, you'll see something like this:
mov edi, OFFSET FLAT:std::cout+8
call std::basic_ios<char, std::char_traits<char> >::operator void*() const
cmp rax, OFFSET FLAT:.LC0
setb al
test al, al
Which makes it clear that operator void*
is the culprit.
Contrary to what Bill Lynch said, I'm able to reproduce it with —std=c++11
on Compiler Explorer. However, it does appear to be an implementation defect, since C++11 should have replaced operator void*
with operator bool
on basic_ios
.
回答2:
This is only valid before C++11.
You're basically doing: ((void *) std::cout) < ((char *) "test")
来源:https://stackoverflow.com/questions/29723423/why-does-instead-of-in-stream-output-still-compile