c++ exam question im stuck on...

Nappy

QUESTION 4

Consider the function:

int apply(string& line,const string& str) {
int c = 0;
int pos = line.find(str);
while (pos > 0) {
c++;
pos = line.find(str,pos);
}
return c;
}

What will the following code output?

string s = "* ** *** ****";

cout << apply(s,"**") << endl;


Anyone have any idea how to answer this, I failed exam miserably and need help...

Cheers

Nappy

padraig_f

the reference for the string.find() function might help: http://www.cplusplus.com/reference/string/string/find/

the first thing you need to answer is, what's the value of 'pos' after the first call to line.find() ?

int pos = line.find(str);

(incidentally it looks to me like there's a bug in that code, but it shouldn't prevent you from interpreting the meaning of it)

fasty

What do you think it does? You'll just fail again if we tell you the output. This is simple stuff as long as you have half a brain and just THINK about it!

toiletduck

Nappy wrote: »

Anyone have any idea how to answer this, I failed exam miserably and need help...

We could just give you the answer but it'd be much better for you to try and work your way through it and take a stab at it yourself.

I recommend putting together a simple program which makes use of that function. Step through it line by line in a debugger and see what's happening.

Nappy

Right ok,

I appreciate help and criticism. Am I right in assuming that the function is finding the first occurence of "**" in
"* ** *** ****";
so its position would be 2 from a choice of 0-12. Is that what would be outputted? Am I on the right track?

Webmonkey

It's doing more than that but you are on the right tracks however I do think there is a slight bug in that code but nothing to prevent you from solving it really and understanding what it is doing.

Nappy

Could you elaborate? What is the bug you speeak of and would I have to list the positions of other occurences of "**" like at 5, 6, 9, 10 and 11? And would the output just be the digits??

Webmonkey

Nappy wrote: »

Could you elaborate? What is the bug you speeak of and would I have to list the positions of other occurences of "**" like at 5, 6, 9, 10 and 11? And would the output just be the digits??

Well you should see from looking at it that the main thing in that that's coming back is the C. What is that doing based on the returned value of the position? That loop also has significance. You have to trace this program out on paper and learn what it is doing. There is no point us telling you the answer. To be come a good programmer you will have to come up with this code, not mind read and understand it so you are better off trying this yourself.

From what I can see

pos = line.find(str,pos);

should be

pos = line.find(str,pos+1) ;

Nappy

while (pos > 0)

On this line hoe is pos ever going to be less than 0, cant it only be between 0-12, wouldnt the loop go on infinitely if the condition is never broken.

Webmonkey

Nappy wrote: »

On this line hoe is pos ever going to be less than 0, cant it only be between 0-12, wouldnt the loop go on infinitely if the condition is never broken.

Well if you cared to take a look at the link padraig_f posted above you'd realise that the find command returns a npos ( static const size_t npos = -1; ) when no occurence of the string is found so yes indeed it can be less than 0. Why else would we be checking this condition.

If it never went below 0, then we would end up in an infinite loop unless we broke out of it some how. Funnily enough that code does go into an infinite loop I think but thats because your lecturer forgot the + 1 to move onto next occurence. There you have it, a little hint dropped in.

wobbles-grogan

Im guessin your programming exam is over at this stage?

Anyway, the above comments put you in the right direction, and you were already on it when you started.

In the spoiler is the explanation of this.
I put it in the spoiler cos you really should figure this out is you want to move forward.

Spoiler:

    while(pos>0){
        c++;
        pos = line.find(str,pos);
    }

this loop simply 'counts' the number of occurences of the 'str' in the 'line'
the count is stored in the variable 'c'.

every time the string "**" is found in line, c is incremented and
the 'pos' variable is set to the position of the occurence of "**" in the line. So when its on the last occurence of "**", the next time it loops through 'pos' will be set to "0", terminating the loop.

The funtion then returns 'c'.

Hope this clears things up for ya