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What will be printed as the result of the operation below:

#include<..>

int x;
int modifyvalue()
{
return(x+=10);
}
int changevalue(int x)
{
return(x+=1);
}
void main()
{
int x=10;
x++;
changevalue(x);
x++;
modifyvalue();
printf("First output:%d\n",x);
x++;
changevalue(x);
printf("Second output:%d\n",x);
modifyvalue();
printf("Third output:%d\n",x);
}

Answer Posted / selloorhari

The Output will be:

First output : 12
Second output : 13
Third output : 14

for changevalue() function:
Even though the value of x(11) is sent to
changevalue() and it returns x(12), no variable is assigned
to capture that 12. So, in main() function, x remains as 11(
not as 12 ) . then it gets incremented and prints the value
12...

And, the Same story for other functions also.....

Is This Answer Correct ?    2 Yes 16 No



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