What is the output of the following program:
void main( )
{
int i = 2, j = 3, k, l ;
float a, b ;
k = i / j * j ;
l = j / i * i ;
a = i / j * j ;
b = j / i * i ;
printf( "%d %d %f %f", k, l, a, b ) ;
}

• Output : 0, 2, 0.000000, 2.000000
  
  k = i/j => 2/3 = 0
  i/j*j = 0*3 = 0
  l = j/i => 3/2 = 1
  j/i*i = 1*2 = 2
  a = i/j => 2/3 = 0.000000 [since a is a float variable, it store the decimal value]
  i/j*j = 0.000000*0.000000 = 0.000000 [ Ibn programming, a smaller no. cant divide a larger no.]
  
  b = j/i => 3/2 => 1.000000
  j/i*i = 1.000000 * 2 = 2.000000
  
  
• 10 years agoHelpfull: Yes(29) No(8)
• As here there is no brackets so first precedence is given to multiplication and then division.
  K = 2/3*3 => 2/9 => 0 (because k is int)
  l = 3/2*2 => 2/4 => 0 (because l is int)
  a = 2/3*3 => 2/9 => 0.222223 (because a is float)
  b = 3/2*2 => 3/4 => 0.75 (because b is float)
• 9 years agoHelpfull: Yes(8) No(5)
• 0 2 0.000000 2.000000
• 10 years agoHelpfull: Yes(1) No(2)
• 0 2 0.0 2.0
• 9 years agoHelpfull: Yes(1) No(1)
• 0, 2, 0.00000, 2.00000
• 9 years agoHelpfull: Yes(0) No(0)
• k=0,l=0,a=0.6,b=0.7
• 9 years agoHelpfull: Yes(0) No(2)
• 2 3 2.0 3.0
• 6 years agoHelpfull: Yes(0) No(0)
• k=0
  l=2
  a=0.0
  b=2.0
• 3 years agoHelpfull: Yes(0) No(1)