Write a program that reads names (from an input file) in the following format (y

Question

Write a program that reads names (from an input file) in the following format (you can assume names are no more than 30 characters if you are using C-string):

First_name Middle_name/Middle_initial Last_name

The program then outputs (to a file) the name in the following format (a period must be place right after the Middle_initial. The first character of First/Middle/Last name must be capitalized):

Last_name, First_name Middle_initial.

The input file name must be "infile" and the output file name must be "outfile". There are unknown number of persons in input files, however, each person's name occupies a line. And each person must have First, Middle/Middle Initial, and Last in the format described above.

Example:
Suppose the input file consists of (special consideration is required for the case "Mary J. Horn"):
Mary J. Horn
John Jerry Doe
Edward David Cain

Then the output file should contain:
Horn, Mary J.
Doe, John J.
Cain, Edward D.

Explanation / Answer

please rate - thanks

be sure when you save the input file you put the name in quotes, since you want no extension

#include <iostream>
#include <fstream>
using namespace std;
int main()
{ifstream in;
ofstream out;
char input[30],first[20],middle[20],last[20];
int i=0,j=0,k=0;
bool midfound;
in.open("infile");
if(in.fail())
     {cout<<"input file did not open please check it ";
     system("pause");
     return 1;
     }
out.open("outfile");
in.getline(input,30);
while(in)
{
i=0;
while(input[i]!=' ')
   {first[i]=input[i];
   i++;
   }
first[i++]='';
for(j=i;j<30;j++)
    if(input[j]==' ')
        {middle[0]=input[i];
         middle[1]='.';
         middle[2]='';
          i=j+1;
         midfound=true;
          j=90;
          }
      else if(input[j]=='')
         {midfound=false;
          j=30;
          }
j=0;
while(input[i]!='')
     {last[j++]=input[i++];
     }
last[j++]='';
out<<last<<", "<<first<<" "<<middle<<endl;

in.getline(input,30);
}
in.close();
out.close();

return 0;
}

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