Typo with “cout < myint”. Why does it work?

问题

I have this code and I searched for hours why it fails to print my income

int const income = 0;
std::cout << "I'm sorry, your income is: " < income;

Until I found I missed to write << but wrote <. Why doesn't the compiler detect this and error out? I'm not sure why comparing cout makes sense?

回答1:

integral constant 0 is also a null pointer constant - it can be compared to the result of ostream's operator void *. Note that it'll fail if the constant has any value but 0.

回答2:

The prototypes of the < operator are like :

    bool T::operator <(const T& b) const;

So I guess the compiler transtype the argument as the type of this instance. Did you enabled all the warnings like -Wall

回答3:

It does compile with g++ 4.4.3

#include  <iostream>

int main (void)
{
   int const income = 0;
   std::cout << "I'm sorry, your income is: " < income;
}

However, when running it with -Wall (good practice!), I got a funny message:

:~/stack$ g++ test.cpp -o temp
:~/stack$ g++ -Wall test.cpp -o temp
test.cpp: In function 'int main()':
test.cpp:5: warning: right-hand operand of comma has no effect

No clue what it actually does (or tries to do)...

回答4:

When I compile this code using GCC 4.3.4, I see a warning:

prog.cpp: In function ‘int main()’:
prog.cpp:6: warning: right-hand operand of comma has no effect

...though why it's a warning rather than an error, I don't know.

EDIT: In fact, I don't know which comma it's referring to either, because this code:

int const income = 0;
std::cout << "I'm sorry your income is: " < income;

...generates the same warning (see here).

来源：https://stackoverflow.com/questions/5665221/typo-with-cout-myint-why-does-it-work

标签

c++

iostream