[Bug c++/55189] New: g++ compiler does not report missing return on function with return type
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55189
Bug #: 55189
Summary: g++ compiler does not report missing return on
function with return type
Classification: Unclassified
Product: gcc
Version: 4.6.2
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
AssignedTo: unassig...@gcc.gnu.org
ReportedBy: meanar...@gmail.com
Compiled with MinGW port of g++, compiles without errors or warnings:
g++ bug.cpp -o bug.exe
Minimal sample bug.cpp:
#include
using namespace std;
class Example
{
protected:
int m_x;
public:
Example()
{
m_x = 0;
}
Example( const Example &ref )
{
m_x = ref.m_x;
}
~Example()
{
}
int get()
{
return m_x;
}
void set( int x )
{
m_x = x;
}
Example &operator =( Example ref )
{
m_x = ref.m_x;
return *this;
}
};
Example func( Example p )
{
Example m;
m.set( p.get() );
// ! COMPILER DOES NOT DETECT ABSENCE OF FUNCTION RETURN !
}
int main()
{
Example m;
m.set( 7 );
Example k;
k = func( m );
// ! CONSEQUENTLY RESULT IS RANDOM !
cout << k.get() << endl;
return 0;
}
[Bug c++/55189] g++ compiler does not report missing return on function with return type
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55189
--- Comment #2 from meanone 2012-11-03 11:11:15
UTC ---
(In reply to comment #1)
> (In reply to comment #0)
> > Compiled with MinGW port of g++, compiles without errors or warnings:
>
> It does warn if you ask it to: -Wall
I can understand that this is ok behavior for C, but to have to
EXPLICITLY ASK typesafe language like C++ to inform me if function
WITH RETURN TYPE does not return anything ?
If this is not bug why shouldn't we make
int n;
n =;
printf("%d",n);
to be legal code, compiler should simply assign whatever it wants. Right ?
Could anyone explain what for should missing return on typical C++ function
be useful at all ?
Not to mention probability that it has been forgotten by mistake.
-Wreturn-type should be set by default.
