<quiz display="simple"> {What will be the output?? <syntaxhighlight lang="c">
int main() {
GATECSE 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">
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">
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">
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.
</quiz>
<quiz display="simple"> {What will be the output?? <syntaxhighlight lang="c">
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">
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">
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">
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.
</quiz>