need help with coding ,

[code builder]

example : The input of argv[2] is declared as base if base argument is b2
how do I check that the number is in the range of 2 - 16.

argument 2 must contain the b i don't have the option of taking it out.

with my coding so far "X" even for range to 2-16 i believe it's because of the b. How can i just check the number ? in this string.




#include <cstdlib>
#include <iostream>
#include <string>

using namespace std;

int main(int argc, char *argv[])
{
if(argc=3)
{
int num1 = atoi(argv[1]);
char base = atoi(argv[2]);


if( base < 2 || base > 16 )
{cout << "X" << endl;

const string digits("0123456789ABCDEF");
bool is_neg = num1 <0 ;
string result;
}

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

[doug4]

First of all, next time you post your code, please use the code tags. They show up as "<>" in the Format box to the right.

As far as your code. In the line:

if(argc=3)

You are actually assigning 3 to arg. This will always return true, so you will always execute the following lines. You want the comparison operator "==".

In the line:
char base = atoi(argv[2]);

You are converting argv[2] trying to an integer value and placing it into a char variable. This is legal, but it's not what you want.

After you output "X", you make a bunch of declarations/assignments, and do nothing with them, so I don't know what you are trying to do with them.

To answer your specific question, you must take argv[2] and treat it as a char*. After you verify that the first character is 'b', then start at the second character and convert it to an integer. Something like:

char* bVar = argv[2];
int bVal;
if (bVar[0] == 'b')
{
    bVal = atoi(bVar + 1);
    cout << "bVal = " << bVal << endl;
}
else
{...}

(Before you are done you should also verify that all characters after the first are digits, but for now just assume the format is correct.)