changing my switch statement

mgm11 (18)
i have this code involving binary and hex, which i have written with the help of a few people. however my friend gave me this switch statement(which i have bolded below), and i was curious to know if there was a simpler way to write it, or make it shorter. it should be noted, i'm willing to learn new techniques to acomplich this.


#include <iostream>
#include <cstdlib>
#include <time.h>
#include <sys/time.h>
#include <errno.h> // not really needed, but keep it here, just in case.
#include <math.h> // keep it here, just in case.

using namespace std;

int num;
char* end;

string newbase(int base, int number)
{
char converter[] = "0123456789ABCDEF";
string temp = "";

for (number; number != 0;)
{
int digit = number % base;
number /= base;


temp.push_back(converter[digit]);

}
return temp;
}


int main(int argc, char *argv[])
{

if (argc == 1) // no parameters on command line, other than the name of the executing file.
{
cout << "3333435,s3333435@student.rmit.edu.au,Matthew_Merigan" << endl ;
system("pause");
return(0) ;
}

//--- START YOUR CODE HERE.


//Parameter Check
if (argc == 2 || argc > 3)
{
cout << "P" << endl;
system("pause");
return 0;
}

if (argc == 3)
{

//comma
string comma(argv[1]);

string::size_type i = comma.find(',', 0);
while (i != string::npos)
{
comma.erase(i,1);
i = comma.find(',', i); // this is so that you don't have to search the entire string again
}


//input
double num = strtod(comma.c_str(),&end);
if(*end != '\0')
{
cout << "X" << endl;
system("pause");
return 0;
}

//range 1
if(num > 1000000 || num < -1000000)
{
cout << "R" << endl;
system("pause");
return 0;
}
//decimal

double dec = (int)num;
if(dec != num)
{
cout << "X" << endl;
system("pause");
return 0;
}

//argv2 for switch
if (argv[2][0] != 'b' || strlen(argv[2]) > 3)
{
cout << "V" << endl;
system("pause");
return 0;
}

if (strlen(argv[2])== 3 && argv[2][1] != '1')
{
cout << "V" << endl;
system("pause");
return 0;
}

//range
if (num > 65535 || num <= 0)
{
cout << "R" << endl;
system("pause");
return 0;
}

int convert = (int)num;

//switch
switch(argv[2][1])
{
case '2': cout << newbase(2, convert) << endl;
break;
case '3': cout << newbase(3, convert) << endl;
break;
case '4': cout << newbase(4, convert) << endl;
break;
case '5': cout << newbase(5, convert) << endl;
break;
case '6': cout << newbase(6, convert) << endl;
break;
case '7': cout << newbase(7, convert) << endl;
break;
case '8': cout << newbase(8, convert) << endl;
break;
case '9': cout << newbase(9, convert) << endl;
break;

case '1':

switch(argv[2][2])
{
case '0': cout << newbase(10, convert) << endl;
break;
case '1': cout << newbase(11, convert) << endl;
break;
case '2': cout << newbase(12, convert) << endl;
break;
case '3': cout << newbase(13, convert) << endl;
break;
case '4': cout << newbase(14, convert) << endl;
break;
case '5': cout << newbase(15, convert) << endl;
break;
case '6': cout << newbase(16, convert) << endl;
break;

default: cout << "V-3" << endl; break;
}

break;

default: cout << "V-4" << endl; break;
}
system("pause");
return 0;
}
system("pause");
return (0);
}
webJose (2948)
There is a super short version to convert a number to a different base. Without delivering the code, I can tell you that:

a. The base-N logarithm of the number + 1 is the number of digits the number has when converted to base N.
b. The simplest algorithm is a recursive one.

That should get you going, I hope.
mgm11 (18)
I’m sorry you’l have to forgive me, it’s my first semester doing computer coding, and I don’t quite understand where you are coming from. do you mean make something like an if statement, with the switch inside of it? something similar to this?

if(switch 13;i++)
{
//etc
}
Caligulaminus (201)
Regarding your switch():

How about:

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int base = atoi(argv[2]);
if(base > 1 && base < 17)
{
    cout << newbase(base, convert);
}
else
{
    // Error
}


And please use code tags.
mgm11 (18)
Caligulaminus, how would that help, i mean to my understanding that code, would check to make sure base is between the right range, but it doesn't output the binary or hex, like the switch does. i also replaced it with the switch, but got no output.
Caligulaminus (201)

it doesn't output

What do you think my cout << newbase(base, convert); does?
mgm11 (18)
k, but i stil found that it didn't output the binary or hex output. this is what i did

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[output]

#include <iostream>
#include <cstdlib>
#include <time.h>
#include <sys/time.h>
#include <errno.h> // not really needed, but keep it here, just in case.
#include <math.h> // keep it here, just in case.

using namespace std;

int     num;
char*       end;

string newbase(int base, int number)
{
    char converter[] = "0123456789ABCDEF";
    string temp = "";
   
    for (number; number != 0;)
    {   
           int digit = number % base;
           number /= base;
       

temp.push_back(converter[digit]);

    }
return temp;    
}


int main(int argc, char *argv[])
{
   
    if (argc == 1)  // no parameters on command line, other than the name of the executing file.
    {
           cout << "3333435,s3333435@student.rmit.edu.au,Matthew_Merigan" << endl ;
           system("pause");
    return(0) ;
    }  

//--- START YOUR CODE HERE.
   
   
//Parameter Check
    if (argc == 2 || argc > 3)
    {
           cout << "P" << endl;
           system("pause");
    return 0;    
    }

    if (argc == 3)
    {
   
//comma
string comma(argv[1]);

string::size_type i = comma.find(',', 0);
while (i != string::npos)
{
comma.erase(i,1);
i = comma.find(',', i); // this is so that you don't have to search the entire string again
}

       
//input   
       double num = strtod(comma.c_str(),&end);
       if(*end != '\0')
       {
               cout << "X" << endl;
               system("pause");
       return 0;
       }
        
//range 1
       if(num > 1000000 || num < -1000000)
       {
               cout << "R" << endl;
               system("pause");
       return 0;
       }
//decimal
       
       double dec = (int)num;
       if(dec != num)
       {
               cout << "X" << endl;
               system("pause");
       return 0;    
       }
       
//argv2 for switch
       if (argv[2][0] != 'b' || strlen(argv[2]) > 3)
       {
               cout << "V" << endl;
               system("pause");
       return 0;
       }

       if (strlen(argv[2])== 3 && argv[2][1] != '1')
       {
               cout << "V" << endl;
               system("pause");
       return 0;
       }

//range
       if (num > 65535 || num <= 0)
       {
               cout << "R" << endl;
               system("pause");
       return 0;
       }
       
       int convert = (int)num;
       
int base = atoi(argv[2]);
if(base > 1 && base < 17)
{
    cout << newbase(base, convert)<<endl;
}
else
{
    cout<<"M"<<endl;
}

    }
system("pause");
return (0);
}

[/output]
Last edited on
Caligulaminus (201)
Well, I don't know how you are calling your Program.

If you have a 'b' prefixed to your base it should probably be
int base = atoi(argv[2]+1); // +1 to skip the 'b'
And you need to
#include <string>
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