Converting 32 bit integer to printable 8 bit Character - c++

Question

I want to convert a series of 32-bit integer values into a sequence of printable 8-bit character values. Mapping the 32-bit integers to printable 8-bit character values should result in a clear ASCII art image.
I can convert Integer to ASCII:
#include <iostream>
using namespace std;
int main() {
char ascii;
int numeric;
cout << "Enter Number ";
cin >> numeric;
cout << "The ascii value of " << numeric << " is " << (char) numeric<<"\n\n"<<endl;
return 0;
}
Also I need to open the text file that my numbers are saved into:
// reading a text file
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main () {
string line;
ifstream myfile ("1.txt");
if (myfile.is_open())
{
while ( myfile.good() )
{
getline (myfile,line);
cout << line << endl;
}
myfile.close();
}
else
cout << "Unable to open file";
return 0;
}
but my problem is , I can not open this " Text " file and print the ASCII on the screen and also print a copy of that in a " Output.txt "
Inside of my Text file is just :
757935403 544999979 175906848 538976380
757795452 170601773 170601727
That after converting to ASCII needs to look like this :
represents the ASCII art picture
+---+
| |
| |
+---+
and have this also in my output.txt.
Please advise if you know how can I write this program.

Answer 1

First of all, you cannot convert a 32 bit integer to 8 bit ascii without losing information. As far as I guess, you should extract 4 ascii chars from a 32 bit integer.
If your input file is non-binary (which means integer values are human-readable/seperated by some delimeter), first thing you should do is create another file/stream and write these values to the new file/stream but now in binary mode (In this mode, there will be no delimiter and resulting file/stream will not be human readable).
Now read chars one by one(open file with binary mode) from this new file/stream, and write it to your final output file using non-binary mode.
IF YOU WANT TO DO IT WITHOUT SEVERAL FILE INOUTS,
Read all your integer values in an array, then point the starting memory location with a char pointer, then write one by one the contents of this char array.
int* myIntArray; //keep the size of it somewhere
char* myCharArray =(char*)myIntArray; // size for myCharArray is 4 times of the myIntArray

Answer 2

Having converted those numbers into hex, you get this
2d2d2d2b 207c0a2b 0a7c2020 2020207C
- - - + | lf+ lf| |
etc etc
so basically for some reason the input file contains the characters to output stored as integers. Which is completely endian unsafe.
Your least worst bet it to read in each integer, cast it to an array of chars and output those 4 chars.
If you're using unix, I'd suggest using 'tee' to send your output to 2 files if you can, otherwise output once to stdout, then output again to whatever file handle you've opened for Output.txt.