Cannot Print All Integer Values - c++

Question

I would like to print integer values in a file by reading it.
The code:
int temp;
char* trainname;
trainname="dfg.txt";
ifstream trainfile;
trainfile.open(trainname);
if(!trainfile){
cout<<"Cannot open file!"<<'\n';
exit(1);
}
while(trainfile >> temp)
cout << temp << " ";
trainfile.close();
dfg.txt: 1 2 we er rf 5
output: 1 2
The problem is that it does not print 5.

Answer 1

Read to a temporary string first and then use std::stoi to try to parse an integer from it, and if it succeeds, output it:
std::string temp;
while(trainfile >> temp) {
try {
std::cout << std::stoi(temp) << " ";
}
catch(const std::invalid_argument&) {
// not a valid number
}
}

Answer 2

while(trainfile >> temp)
cout << temp << " ";
The above sets the failbit on trainfile on encountering any character that isn't whitespace or a digit. That terminates the loop. This is one reason I tend not to use formatted I/O that can fail on a input stream. I find it better to read text as text (not numbers) and then process the string that was just read. For example, see zenith's answer.
If you insist on doing everything from the input stream, you'll need an outer loop that clears the stream's failbit. For example,
while (! std::cin.eof())
{
while (std::cin >> temp)
{
std::cout << temp << " ";
}
std::cin.clear();
std::cin.ignore();
}
Given an input file containing 1 2 we er rf 5, the above will print 1 2 5. If the input file contains 1 2 abc345def 6, the above will print 1 2 345 6. Note that zenith's approach will print 1 2 6. Whether that 345 sandwiched between abc and def counts as an integer is up to you.
I'd recommend using zenith's solution over mine.
Update:
The above interprets abc345def as representing the integer 345. Both Zenith's solution and the above interpret 345def as representing the integer 345. To me, both abc345def and 345def should be rejected as representing an integer. So should 6.1 , but there's nothing wrong with 0x abc345def. There's nice tool in the C standard library, strtol, that nicely parses integers. It also indicates what made the parse stop. For a valid integer, it should stop at the end of the input string. With that,
#include <iostream>
#include < fstream>
#include <string>
#include <cstdlib>
int main ()
{
std::ifstream trainfile("dfg.txt");
if (!trainfile)
{
std::cerr << "Cannot open file!\n";
exit(1);
}
std::string s;
while(trainfile >> s)
{
char* end;
long num = std::strtol (s.data(), &end, 0);
if (!*end)
{
std::cout << num << " ";
}
}
trainfile.close();
std::cout << "\n";
}

Answer 3

string temp;
if( '0' <= temp[0] && temp[0]<='9' )
cout << temp << " ";
it will work i suppose.

Answer 4

Here is another way you can consider-
#include <iostream>
#include <sstream>
#include <fstream>
using namespace std;
int main()
{
ifstream trainname("dfg.txt",ios::in);
string temp;
getline(trainname,temp);
stringstream str;
str<<temp;
int extract_int;
while(getline(str, temp,' '))
{
if(stringstream(temp)>>extract_int)
cout<<extract_int<<" ";
}
return 0;
}
Or according to David Hammen's answer, you can solve it the following way-
#include <iostream>
#include <sstream>
#include <fstream>
using namespace std;
int main()
{
int temp;
char* trainname;
trainname="dfg.txt";
ifstream trainfile;
trainfile.open(trainname);
if(!trainfile){
cout<<"Cannot open file!"<<'\n';
exit(1);
}
while (!trainfile.eof())
{
while (trainfile>>temp)
cout<<temp<< " ";
trainfile.clear();
trainfile.ignore();
}
return 0;
}