Populating Vector from string Help

Below is a snippet of code that works for me. Currenlty it populates
my vector with one digit per element. I would like to modify this
code to populate each element of the vector with 3 digits from the
string.

(i.e. 179,534,672,198 would be stored with block [0] = 198, block [1]
= 672, block [2] = 534, and block [3] = 179)

vector<intvOne, vTwo;
string one, two;
size_t r;

cout << "\n\n\t Please enter the first integer:\t";
cin>>one;

r = one.length();

for (int x = 0; x < r; x++){
vOne.push_back(int(one[x] - '0'));
}

re: Populating Vector from string Help

yogi_bear_79 wrote:
This example is poorly chosen since it hides an important detail. How do you
want to store "12345"? Do you want 12.345 or 123.45?

[snip]


Best

Kai-Uwe Bux

re: Populating Vector from string Help

On Mar 16, 9:49*pm, Kai-Uwe Bux <jkherci...@gmx.netwrote:I would want it to be [012] {345]

re: Populating Vector from string Help

On Mar 16, 6:56 pm, yogi_bear_79 <yogi_bear...@yahoo.comwrote:Accosding to your first post you mean:
[345]
[012]

But will the input include ',' or were they just to help explain?
ie. Is the input going to be a number or a real string.

re: Populating Vector from string Help

On Mar 17, 12:28*am, Martin York <Martin.YorkAma...@gmail.comwrote:just a number, no other characters will be present

re: Populating Vector from string Help

On Mar 17, 12:48 pm, yogi_bear_79 <yogi_bear...@yahoo.comwrote:It's a weird requirement, but:

std::string::const_iterator end = myString.end() ;
std::string::const_iterator begin = myString.begin() ;
while ( end - begin >= 3 ) {
std::istringstream s( std::string( begin, begin + 3 ) ) ;
int i ;
s >i ;
// Error handling goes here...
myVect.push_back( i ) ;
begin += 3 ;
}

In general, when you need to break a string (or any container
supporting random access iterators) up into smaller blocks, you
can create an instance using two iterators, which delimit the
smaller block. And any time you have to convert from or to
strings, istringstream or ostringstream do the trick.

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