Question

bool foo( int* p )
{
  return false;
}

int* bar (  )
{
  int* p;
  return p;
}

int main() {

  if ( int* p = bar() && foo( p ) )
  //if ( ((int* p = bar()) != NULL ) && (foo( p )) ) // another variant
  {

  }

  return 0;
}

I get error: cannot convert ‘bool’ to ‘int*’ in initialization with GCC 4.3.4 and VS 2008. If there is only if ( int* p = bar()) everything works fine.

I know i can initialize p before if-statement, but i’m interested in the variant i shown.

Answer 1

The syntax of an if requires a condition in the parentheses. A
condition can be either an expression, or a declaration (of a
type with an implicit conversion to bool); it cannot be a combination
of these. In the case of

if ( int* p = bar() && foo( p ) )

, the condition is “int* p = bar() && foo( p )“, which is a simple
declaration of a type int* initialized with bar() && foo( p ); since
the type of bar() && foo( p ) isn’t int*, and can’t be implicitly
converted into an int*, you have an error.

If you write:

if ( (int* p = bar()) != NULL )

, the opening parentheses mean that this cannot be a declaration (simple
or no), and int* p = bar() isn’t a legal (sub-)expression.

The critical point to remember is that you can have a declaration, or
you can have an expression, but that you can’t combine them, and the C++
doesn’t allow declarations as part of an expression. (Also: the
declaration must have copy initialization:

if ( int* p = bar() )

, direct initialization is not allowed.)