View source for C Questions on Expressions ← C Questions on Expressions


<quiz display="simple">
{'''What will be the output??'''
<syntaxhighlight lang="c">
#include <stdio.h>
#include <string.h>

int main()
{
   int x = 8, y = 1;
   switch(x--, y++)
   {
       case 1: x*=8;
       case 2: y*= x/=2;
       case 3:
       case 4: y--;
       default: x+=5;
   }
   printf("%d %d", x, y);
}

</syntaxhighlight>
|type="()"
/}
-64 2
-64 1
+33 55
-33 56
|| (x--, y++) will return the value of y which is 1, as comma operator always returns the right value. Hence, switch case starts with 1. Before starting the switch case, x is decremented and y incremented also. So, in case 1, x is 7 and y is 2. x is changed to 56 in case 1. Because of no break, all cases are evaluated here. So, in case 2, x becomes 28 and y becomes 56. In case 3 nothing happens. In case 4, y becomes 55 and finally in default case x becomes 33. 
 
{'''What will be the output??'''
<syntaxhighlight lang="c">
#include <stdio.h>
#include <string.h>

int main()
{
  int i = 3, j = 6, k;
  k = (i++ * j, ++i * j);
  printf("%d",k);
}

</syntaxhighlight>
|type="()"
/}
-Undefined behavior
-Compiler dependent
+30
-24
||comma operator always returns the right operand and it also forms a sequence point (comma in function parameter list is actually just a separator and not comma operator thus does not form a sequence point). Hence post increment of i is done before comma operator starts and then pre increment is done and finally i (5) is multiplied by j (6) giving 30.

</quiz>