c++ - Union float and int -

i'm little bit confused. during development of 1 function based on predefined parameters, pass sprintf function exact parameters needed based on type, found strange behaviour ( "this %f %d example", typefloat, typeint ).

please take @ following stripped working code:

struct param {  enum { typeint,    typefloat   } paramtype;  union {     float f;     int i;  }; };  int _tmain(int argc, _tchar* argv[]) {  param p;  p.paramtype = param::typeint;  p.i = -10;   char chout[256];  printf( "number %d\n", p.paramtype == param::typeint ? p.i : p.f );  printf( "number %f\n", p.paramtype == param::typeint ? p.i : p.f );  return 0; } 

my expected output printf( "number %d\n", p.paramtype == param::typeint ? p.i : p.f );

number -10 

but printed

number 0 

i have put breakpoint after initialization of param p, , although p.paramtype defined typeint, actual output if -10.0000. checking p.f gives undefined value expected, , p.i shows -10 expected. p.paramtype == param::typeint ? p.i : p.f in watch window evaluates -10.000000.

i added second printf prints float , output is

number 0 number -10.000000 

so why happens? possible bug in visual studio (i use vs2012)?

update:

std::cout<< (p.paramtype == param::typeint ? p.i : p.f); 

gives correct value of -10.

this because resulting type of expression p.paramtype == param::typeint ? p.i : p.f float (only value different depending on paramtype), first format string expects integer.

the following should work expect:

printf("number %d\n", (int)(p.paramtype == param::typeint ? p.i : p.f)); 

the cout version gives expected output because type of expression being inserted (float) deduced automatically, , formats float. same second printf.