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Microsoft KB Archive/104243

From BetaArchive Wiki

Article ID: 104243

Article Last Modified on 7/5/2005


  • Microsoft C/C++ Professional Development System 7.0
  • Microsoft Visual C++ 1.0 Professional Edition
  • Microsoft Visual C++ 1.5 Professional Edition
  • Microsoft Visual C++ 1.51
  • Microsoft Visual C++ 1.52 Professional Edition
  • Microsoft Visual C++ 1.0 Professional Edition
  • Microsoft Visual C++ 2.0 Professional Edition
  • Microsoft Visual C++ 2.1
  • Microsoft Visual C++ 4.0 Standard Edition

This article was previously published under Q104243


The following error

Error C2440 : 'Conversion': cannot convert from 'type1' to 'type2'

is generated when a statement of the form

   TYPE ( identifier ) ( expression );

is used to type cast a function pointer and make a call simultaneously.


This is expected C++ compiler behavior. The sample code below explains the problem in a more understandable way.

If the following three statements taken from the sample code are compiled

   typedef void (* funcptr) (int);


   void f(int){...};


   funcptr (f)(1); // Error C2440

the following error message is generated:

   error C2440: 'initializing' : cannot convert from 'const int' to
   'void (__cdecl *) (int)'

Although the last statement appears to be typecasting the function "f" to funcptr, the compiler thinks it is declaring an object of name "f" of type funcptr and initializing it with the integer constant "1" equivalent to having

   funcptr f = 1;


   void (*f) (int) = 1;

which makes the error message much more understandable.


The way to the achieve the desired typecast is to add another set of parenthesis, for example


which resolves the confusion for the compiler, which thus treats the typecast in the way it should.

NOTE: Using a C++ style cast, such as (funcptr(f))(1), produces a syntax error. You must use C-style casting.

Another method is to have a pointer explicitly bind a new name to the function, for example

   void (*f1)(int) = f;

and use:

   f1(1);     // OK


The sample below demonstrates the error and workarounds.

Sample Code

typedef void (*funcptr) (int);

void f(int);

void main()
    ((void (*) (int)) f)(1);
    funcptr (f) (1);             // Gives error C2440
    void (*f1) (int) = f;

Additional query words: 1.00 1.50 2.00 2.10 4.00 7.00 8.00 8.00c 9.00 9.10

Keywords: kbprb KB104243