String Verification (Checks if string has certain characters or not)

[MasterTroll]

So, I'm trying to make a program to convert numbers between bases and I'm stuck on number verification, where it checks the string for only the digits 0-9 and/or letters A-F. But, when I enter something like "19s", it just skips telling me the error message I want. Any solutions for a rewrite?

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

using namespace std;

int main() {
    cout << "Number Base Converter Version 1.00" << endl;
    cout << "Enter a number you want to convert: ";
    string number;
    cin >> number;
    int i = 0, status = 0;
    while (status = 0) {
        for (int i = 0; i < number.length(); i++) {
            if (number.at(i) != '0' || '1' || '2' || '3' || '4' || '5' || '6' || '7' || '8' || '9' || 'A' || 'a' || 'B' || 'b' || 'C' || 'c' || 'D' || 'd' || 'E' || 'e' || 'F') {
                cout << "The number you enter must only contain the digits 0-9 and/or letters A-F." << endl;
                cout << "Enter a number you want to convert: ";
                cin >> number;
                status = 0;
            } else {
                status = 1;
            }
        }
    }

[cire]

if ( x != '0' || '2' )

evaluates to

if ( (x!='0') || '2)'

In the case where x is '0':

if ( ('0'!='0') || '2' )

which becomes:

if ( false || '2' )

and since '2' is non-zero and is always true, the control expression is also always true.

The correct way to write your if would be :

char c = number.at(i) ;
if ( c != '0' && c != '1' && c != '2' && c != '3' /* ... */ )

You might want to try something along the lines of:

#include <algorithm>

bool is_hex_digit(char c )
{
	static const char digits[] = "0123456789ABCDEFabcdef" ;
	static const char* digits_end = digits + (sizeof(digits) / sizeof(digits[0])) ;

	return digits_end != std::find(digits, digits_end, c) ;
}

Or a switch construct may be more to your taste.