What is the output of the following program?
#include
void main()
{
printf("%d",sizeof(5.2));
}

• output : 8
  
  Since 5.2 is double value, no. of bytes allocated to a double value is 8...
  Hence it ll print 8...
• 10 years agoHelpfull: Yes(23) No(1)
• answer:8
  By default compiler consider 5.2 as a double
  And We know that size of double is 8.
  If we want to define 5.2 as a float we can declare 5.2 as 5.2f
• 10 years agoHelpfull: Yes(4) No(0)
• Output is 8.
  bcoz 5.2 is a double type...
  
  if there had been sizeof(5.2f) then the answer would have been 4 bytes
• 9 years agoHelpfull: Yes(3) No(0)
• by turbo compiler and others also 5.2 will be by default considred as long therefore, size will be 8 .
• 10 years agoHelpfull: Yes(2) No(0)
• 2 as it is int
• 10 years agoHelpfull: Yes(1) No(9)
• 4
  sizeof the float is 4
• 10 years agoHelpfull: Yes(1) No(5)
• 5.2 is taken as double type by default so size is 8
• 10 years agoHelpfull: Yes(0) No(0)
• out put:8
  The compiler treats it as double.The size of double is 8 bits... so it will print 8 as output.
• 10 years agoHelpfull: Yes(0) No(0)
• its a float so answer is 8
• 10 years agoHelpfull: Yes(0) No(0)
• 2 bytes
  as %d specifies integer.
• 10 years agoHelpfull: Yes(0) No(2)
• ans : 8
  5.2 represents the double .so ,double size is 8
• 8 years agoHelpfull: Yes(0) No(0)
• 8
  Since every floating point literal is of "double" type.
• 7 years agoHelpfull: Yes(0) No(0)
• 8
  size of double
• 7 years agoHelpfull: Yes(0) No(0)
• this is outpt of 5.2
• 7 years agoHelpfull: Yes(0) No(1)
• it prints 8
• 4 years agoHelpfull: Yes(0) No(0)