Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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| Line 19: | Line 19: | ||
printf("%d %d", x, y); | printf("%d %d", x, y); | ||
} | } | ||
| + | |||
</syntaxhighlight> | </syntaxhighlight> | ||
|type="()" | |type="()" | ||
| Line 32: | Line 33: | ||
#include <stdio.h> | #include <stdio.h> | ||
#include <string.h> | #include <string.h> | ||
| − | |||
int main() | int main() | ||
{ | { | ||
| − | int i = 3, j = 6, k; | + | int i = 3, j = 6, k; |
| − | + | k = (i++ * j, ++i * j); | |
| − | k = (i++ * j, ++i * j); | + | printf("%d",k); |
| − | printf("%d",k); | ||
} | } | ||
<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.
</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.
</quiz>