Difference between revisions of "C Questions on Expressions"

Revision as of 16:39, 19 June 2014

<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>

@@ Line 28: / Line 28: @@
 || (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>