Understanding output while using flags in printf() in c programming [duplicate]

Question

I was studying about formatted input output functions and found out that i could actually use printf() in a variety of ways, so I started trying out how to use flags in printf() . I was trying to understand # flag , it was stated that if I use it with e,E,f format specifier it "forces the written output to contain a decimal point even if no digits would follow ". So I thought what would happen if i did it with integer but with float formating and also + flag .

main()
{
  int  b = 0 , c = 4.53 , d = 3 ;
  float a = 1 ;
  double e = 5.32 ;//please dont mind a,b,c they are remanents from previous testing
  printf("%+#f" , d);    
  return 0 ;
}

Output - -1.#QNAN0

Your Help would be highly appreciated . Thanx in advance for taking your time to help me

David Ranieri · Accepted Answer · 2021-03-13 09:47:57Z

0

You are invoking undefined behaviour, you can not pass an int when printf is expecting a double.

Variadic functions promotes float to double, so this works even if %f expects a double:

float f = 3.14;
printf("%f", f);

It also promotes char and short int to int:

char c = 1;
printf("%d", c);

But can not promote an int to double (as you were expecting), a cast is needed:

printf("%+#f", (double)d);

edited Mar 13, 2021 at 9:47

answered Mar 13, 2021 at 9:26

David Ranieri

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