Read from string into stringstream

Question

When I try to parse whitespace seperated double values from a string, I found this curious behaviour that the string is read out in a cyclic manner.

Here's the program:

stringstream ss;
string s("1 2 3 4");
double t;
list<double> lis;

for(int j=0; j!=10; ++j){
    ss << s;
    ss >> t;
    lis.push_back(t);
}

for(auto e : lis){
    cout << e << " ";
}

Here the output:

1 2 3 41 2 3 41 2 3 41

If I append a trailing space as s= "1 2 3 4 "; I get

1 2 3 4 1 2 3 4 1 2 

Now the questions:
1) If I don't know how many entries are in the string s, how do I read all into the list l?
2) which operator<< am I actually calling in ss << s;? Is it specified to read circularly?
3) Can I do the parsing in a better way?

Thanks already!

---

Here's the fixed code (thanks to timrau):

// declarations as before
ss << s;
while(ss >> t){
   lis.push_back(t);
}
// output as before  

This produces:

 1 2 3 4  

as desired. (Don't forget to clear your stringstream by ss.clear() before treating the next input. ;))

Another useful comment from HeywoodFloyd: One could also use boost/tokenizer to "split" the string, see this post

Answer 1

You can test the return value of >>.

while (ss >> t) {
    lis.push_back(t);
}

It's not specified to read circularly. It's ss << s appending "1 2 3 4" to the end of the stream.

Before the 1st loop:

""

After 1st ss << s:

"1 2 3 4"

After 1st ss >> t:

" 2 3 4"

After 2nd ss << s:

" 2 3 41 2 3 4"

Then it's clear why you get 1 2 3 41 2 3 41 2 3 41 if there is no trailing space in s.

Answer 2

then use s.length() for strings containing unknown number of entries, if you use your approach. Or, as suggested by timrau, just initialize your stringstream once.

stringstream ss;
string s("1 2 3 4 5 6 7 8");
ss << s;
double t;
list<double> lis;
while (ss >> t) {
    lis.push_back(t);
}

for(auto e : lis){
    cout << e << " ";
}

Answer 3

This stackoverflow post includes a boost tokenizer example. You may want to tokenize your string and iterate through it that way. That will solve the no trailing space problem timrau pointed out.