Question

Convert the following C program to C++.

More instructions follow the code.

#include <stdio.h>
#include <stdlib.h>

#define SIZE 5

int main(int argc, char *argv[]) {
   int numerator = 25;
   int denominator = 10;
   int i = 0;

   /*
   You can assume the files opened correctly and the
   correct number of of command-line arguments were
   entered.
   */
   FILE * inPut = fopen(argv[1], "r");
   FILE * outPut = fopen(argv[2], "w");

   float result = (float)numerator/denominator;
   fprintf(outPut,"Result is %.2f\n", result);

   float arr[SIZE];

   for( ; i < SIZE; i++) {
      fscanf(inPut, "%f", &arr[i]);
      fprintf(outPut, "%7.4f\n", arr[i]);
   }

   return 0;
}

Notice this is uses command-line arguments. I have provided an input file called  num.txt that will be used when running the program. The output file is called  out.txt.

Make sure you are using C++ style file I/O (FILE pointers/fopen) as well as regular I/O including the C++ style output formatting (fscanf, fprintf, formatting). Also use the C++ method of casting. The lines above that are bold are the lines that you need to convert to C++. Don't forget to add the necessary C++ statements that precede the  main() function.

Answer 1

Below is the given code in C++:

#include<iostream>
#include<fstream>
#include<iomanip>
#define SIZE 5

using namespace std;

int main(int argc, char *argv[])
{
int numerator = 25;
int denominator = 10;
int i = 0;

// open a file in read mode.
ifstream inPut;
inPut.open(argv[1]);

// open a file in write mode.
ofstream outPut;
outPut.open(argv[2]);

float result = (float)numerator/denominator;
outfile<<fixed<<setprecision(2)<<"Result is "<<result<<endl;
float arr[SIZE];

for( ; i < SIZE; i++) {
infile >> arr[i];
outfile<<fixed<<setprecision(4)<<arr[i]<<endl;
}
return 0;
}

Most of the syntax is same, just that of file handling is different.

NOTE: Install stream library if it is not present.