Programming puzzles · DIGIT-ARC

S

Post some interesting " find the output " type questions or any other sort of tricky programs here in c++ or java....

take the following code:

int i, n = 20;
for (i=0; i<n; i--)
{
cout << "x" << endl;
}

by changing only ONE character in the above code, meaning you cannot change 20 to 31, because you will have changed two characters, you can change 20 to 21, because you only changed the 0, do the following:

find 3 ways to make the above code print x 20 times (by changing only one character or inserting one char).

Tell the output:

float f=1.22;
if(f == 1.22)
printf("True");
else
printf("False");

Go to www.hackerrank.com and knock yourselves out.

sid_gupta said:
Post some interesting " find the output " type questions or any other sort of tricky programs here in c++ or java....

take the following code:

int i, n = 20;
for (i=0; i<n; i--)
{
cout << "x" << endl;
}

by changing only ONE character in the above code, meaning you cannot change 20 to 31, because you will have changed two characters, you can change 20 to 21, because you only changed the 0, do the following:

find 3 ways to make the above code print x 20 times (by changing only one character or inserting one char).

int i, n = 20;
for (i=0; i<n; i--)
{
count << "x" << endl;
}

sid_gupta said:
Post some interesting " find the output " type questions or any other sort of tricky programs here in c++ or java....

take the following code:

int i, n = 20;
for (i=0; i<n; i--)
{
cout << "x" << endl;
}

by changing only ONE character in the above code, meaning you cannot change 20 to 31, because you will have changed two characters, you can change 20 to 21, because you only changed the 0, do the following:

find 3 ways to make the above code print x 20 times (by changing only one character or inserting one char).

if i=0, why are you doing a i--? that will go into an infinite loop.

S

well that is the the question....
you have to figure out 3 diff. ways to make this loop print 20 times by changing only one character in the code.....
hint :- you can also change whitespaces

N

sid_gupta said:
Post some interesting " find the output " type questions or any other sort of tricky programs here in c++ or java....

take the following code:

int i, n = 20;
for (i=0; i<n; i--)
{
cout << "x" << endl;
}

by changing only ONE character in the above code, meaning you cannot change 20 to 31, because you will have changed two characters, you can change 20 to 21, because you only changed the 0, do the following:

find 3 ways to make the above code print x 20 times (by changing only one character or inserting one char).

Here's one way to do it.

PHP:

int i, n = 20; 
for (i=0; i<n; n--)  
{ 
   cout << "x" << endl; 
}

sid_gupta said:
well that is the the question....
you have to figure out 3 diff. ways to make this loop print 20 times by changing only one character in the code.....
hint :- you can also change whitespaces

Here's one solution. You tell the other two:

Code:

 int n = 20;
        for(int i = 0; [B][SIZE=4]-i[/SIZE][/B]<n;i--)
        {
            cout << "x" << endl; 
            
        }

Just add a "-" prefix to i.

N

Nue said:
Here's one way to do it.

PHP:

int i, n = 20; for (i=0; i<n; n--) { cout << "x" << endl; }

Okay so here's another method:

PHP:

int i, n = 20;  
for (i=0; i+n; i--)   
{  
    cout << "x" << endl;  
}

So that's 2 and + 1 vickybat's answer. I guess your puzzle's been solved.