C++ iterator value to variable

Question

I am using an iterator in a C++ code to retrieve records read using sqlite3 statements. I am able to display the contents pointed to by the iterator to screen using cout. How would i assign the value to a simple float or array variable.

typedef vector<vector<string> > Records;
vector< vector<string> >::iterator iter_ii;
vector<string>::iterator iter_jj;

Records records = select_stmt("SELECT density FROM Ftable where PROG=2.0");

  for(iter_ii=records.begin(); iter_ii!=records.end(); iter_ii++)
   {
      for(iter_jj=(*iter_ii).begin(); iter_jj!=(*iter_ii).end(); iter_jj++)
      {
         cout << *iter_jj << endl; //This works fine and data gets displayed!

         //How do i store the data pointed to by *iter_jj in a simple float variable or array?
      }
   }

*iter_jj is basically a string, so you need to convert that to the type you want. Have a look [here](http://stackoverflow.com/questions/1012571/stdstring-to-float-or-double) for a host of methods — StevieG, Sep 18 '13 at 15:32

score 2 · Answer 1 · answered Sep 18 '13 at 15:31

C++ is type-safe, so you need to explicitly convert the string to the desired target type.

For float for example you could use atof:

float f = atof(iter_jj->c_str());

A more convenient alternative is Boost's lexical_cast, which works with the same syntax for all types that support extraction from an std::istream:

float f = boost::lexical_cast<float>(*iter_jj);

Note that both of these can fail in different ways if the contents of the string cannot be converted to a float in any meaningful way.

score 0 · Accepted Answer · answered Sep 18 '13 at 15:32

0

Your real problem is how to convert a string to a float. Here is one solution.

float value;
stringstream ss(*iter_jj);
if (! (ss >> value))
{
    ERROR failed to convert value
}

answered Sep 18 '13 at 15:32

Neil Kirk

21,327
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Thank you very much for the answers! Both the stringstream and atof methods work great! – user2785789 Sep 18 '13 at 15:58

score 0 · Answer 3 · answered Sep 18 '13 at 15:32

0

If you have C++11 compatible compiler:

float x = stof(*iter_jj);

(Obviously x could be a variable outside of the loop).

If you don't have C++11:

stringstream ss(*iter_jj);
float x;
ss >> x;

answered Sep 18 '13 at 15:32

Mats Petersson

126,704
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score 0 · Answer 4 · answered Sep 18 '13 at 15:35

Well since you are working with:

std::vector<std::vector<std::string> > records;

the actual question here is: how to retrieve the specific type of data from std::string object.

The good approach would be constructing and using std::istringstream object for this purpose:

float f;
std::istringstream is(*iter_jj);
if (is >> f)
{
    // handle successful retrieval...
}

just don't forget to #include <sstream> :)

xKommando · Answer 5 · 2013-09-20T02:53:03.930

As for converting * to string, in c++11, you can convert integer/floats to string by calling static method to_string:string str = std::string::to_string(integer/* or float*/);

in c++98, you can write your own to_string:

#include <cstdio>
#include <cstring>
#include <cstdarg>
#include <string>
void format_aux(char* ptr, int size, const char* format, ...) {
    va_list args;
    va_start(args, format);
    vsnprintf(ptr, size, format, args);
    va_end(args);
}

#undef TO_STRING__GEN
#define TO_STRING__GEN(type, arg, size)     \
std::string                                 \
to_string(type val) {                       \
    const int sz = size;                    \
    char buf[sz];                           \
    format_aux(buf, sz, arg, val);          \
    return std::string(buf);                \
}                                           

TO_STRING__GEN(int,           "%d", 4*sizeof(int))
TO_STRING__GEN(unsigned int,  "%u", 4*sizeof(unsigned int))
TO_STRING__GEN(long,          "%ld", 4*sizeof(long))
TO_STRING__GEN(unsigned long, "%lu", 4*sizeof(unsigned long))
TO_STRING__GEN(float,         "%f", (std::numeric_limits<float>::max_exponent10 + 20))
TO_STRING__GEN(double,        "%f", (std::numeric_limits<float>::max_exponent10 + 20))

#undef TO_STRING__GEN

score -1 · Answer 6 · answered Sep 18 '13 at 15:31

*iter_jj is going to give you a std::string. In order to store that as a float, it will need to be a floating point number in string form (e.g. "1.23456") and you will need to call one of the strtof family of functions (http://en.cppreference.com/w/cpp/string/byte/strtof)

C++ iterator value to variable

6 Answers6