CODEWITHSUNDEEP
c++stringstream
user8305079
user8305079

Reputation:

About stringstreams in C++

While studying for interview questions, I came across this piece of code on the internet:

class Solution {
public:
    string complexNumberMultiply(string a, string b) {

        //I initially did it using substring (searching for "+" etc.)
        //But it is super easy using stringstreams

        stringstream aa(a), bb(b), ans;
        int ra, rb, ia, ib;
        char buff;
        aa>>ra>>buff>>ia>>buff;
        bb>>rb>>buff>>ib>>buff;

        ans<<ra*rb-ia*ib<<"+"<<ra*ib+rb*ia<<"i";
        return ans.str();
    }
};

This snippet multiplies two input strings represent complex numbers of the form a+bi. Thus, if the input is 1+1i and 1+1i, then the output generated by this code is 0+2i (because i^2=-1).

I do understand why stringstreams aa and bb have been used and how they work; but I fail to understand the role of char buff. Consider the statement:

aa>>ra>>buff>>ia>>buff;

Here, from the stringstream aa we first read the rational part ra (then buff for a plus "+"?), then the imaginary part ia and then the buff again (maybe for \n?). Is my understanding correct? If I remove the buffers though, it works fine for the inputs like 1+2i but fails where imaginary part is negative like 1+-2i (yeah, not 1-2i).

Kindly let me know if my understanding is correct. Thank you!

Upvotes: 0

Views: 307

Answers (2)

myaut
myaut

Reputation: 11484

Yes, your understanding is correct. operator >> is common for each input stream (ancestor of std::istream) and it uses std::num_get::get for integers.

Accoding to stage 2 logic and positive imaginary part aa >> ra >> rb will run like this:

2 -> accumulated to storage "2" 
     (because 2 matches one of "0123456789abcdefxABCDEFX+-")
+ -> ignored, because "2+" is not valid integer for scanf, so first
     operator >> terminates here
+ -> accumulated to storage "+"
3 -> accumulated to storage "+3"
i -> ignored, because "+3i" is not a valid integer, so we 
     can scan "+3" as integer 3 and put it to rb

Negative imaginary part is broken for second integer:

+ -> accumulated to storage "+"
- -> "+-" is not a valid scanf, so we should terminate second 
     operator >> while putting 0 to ra because "+" is also is not 
     an integer

So by adding explicit reading of single character, you read '+' in 2nd operator >> and can read "3" or "-3" correctly in 3rst call to operator >>

Upvotes: 0

NathanOliver
NathanOliver

Reputation: 180415

You are almost correct. When you have a string like 1+1i you have two valid integers and two valid characters. So aa>>ra reads the first integers into ra leaving you with +1i. Then >>buff reads the character (+) into buff leaving 1i in the stream. Then >>ia reads the next integer and leaves i in the stream. Then >>buff consumes the i left in the stream.

Generally when doing something like this I like to use a more descriptive variable name. I like to use eater a lot as it implies that I'm just eating the input(throwing it away). If I know what the input will look like then even more descriptive names are nice like sign/operator/imaginary_part.

Upvotes: 3

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