Difference between revisions of "C Questions on Expressions"

Latest revision as of 16:16, 6 July 2014

<quiz display="simple"> {What will be the output?? <syntaxhighlight lang="c">

1. include <stdio.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">

1. include <stdio.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.

{What will be the output?? <syntaxhighlight lang="c">

1. include <stdio.h>

int main() {

 int i = 5;
 i = (++i,i++,i) ;
 printf(" i = %d",i);

}

</syntaxhighlight> |type="()" /} -6 -Undefined behaviour +7 -5 ||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 pre-increment of i is done, then post increment and finally i is evaluated to 7. (Between each increment a sequence point is formed by , and hence no undefined behavior)

{What will be the output?? <syntaxhighlight lang="c">

1. include <stdio.h>

int main() {

 int i = 5;
i = (i,++i,++i);
 printf(" i = %d",i);

}

</syntaxhighlight> |type="()" /} -6 +Undefined behaviour -7 -5 ||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 first pre increment of i is done and then again pre increment is done. But between the second pre increment and the assignment operation there is no sequence point (like i = ++i). So, this produces undefined behaviour.

{What will be the output?? <syntaxhighlight lang="c">

1. include <stdio.h>

int main() {

   int x=1,y=2;
   x=((x>0) && (y=6 ? printf("TRUE"): printf("FALSE")));
   printf("\nx = %d y = %d",x, y);

}

</syntaxhighlight> |type="()" /} -True
x = 1 y = 4 -False
x = 0 y = 6 +True
x = 1 y = 4 -False
x = 1 y = 6 ||x=((x>0) && (y=6 ? printf("TRUE"): printf("FALSE")));
After precedence rule
x=((x>0) && (y=(6 ? printf("TRUE"): printf("FALSE"))));
printf returns the number of characters printed. So, here it will print TRUE and returns 4 which is assigned to y and is also returned as the second operand for &&. Since, both the operands of && are non-zero, x is assigned 1.

</quiz>