How to say != 0-9 in c ++ [duplicate]

Answer 1

It's clumsy enough to use that an ad hoc solution may be easier to use, but the standard library actually supports this directly:

#include <locale>
#include <iostream>
#include <iomanip>

int main() {

    char *inputs[] = { 
        "0983",
        "124test"
    };

    std::locale loc(std::locale::classic());

    std::ctype_base::mask m = std::ctype_base::digit;

    for (int i=0; i<2; i++) {
        char const *b = inputs[i];
        char const *e = b + strlen(b);

        std::cout << "Input: " << std::setw(10) << inputs[i] << ":\t";

        if (std::use_facet<std::ctype<char> >(loc).scan_not(m, b, e) == e)
            std::cout << "All Digits\n";
        else
            std::cout << "Non digit\n";
    }
    return 0;
}

If you're using C++11, std::all_of is almost certainly easier to use:

#include <string>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <ctype.h>

int main() {
    std::string inputs[] = {
        "0983",
        "124test"
    };

    std::cout << std::boolalpha;

    for (int i=0; i<2; i++)
        std::cout << std::setw(10) << inputs[i] << "\tAll digits?: "
            << std::all_of(inputs[i].begin(), inputs[i].end(), ::isdigit) << "\n";
    return 0;
}

Answer 2

If the string was "0983" I want it to be true but if it was "124Test" I want it to remain false.

Saying that, an approach would be checking if a character is not a number, then return false, instead of checking every character until the end of the string.

bool b = true;
for(int i = 0; i < string.size(); i++)
{
    if(string.at(i) < '0' || string.at(i) > '9')
    {
        b = false;
        break;
    }
}

Answer 3

Say string[i] < 0 || string[i] > 9.

Make sure you actually mean 0 (the value), and not '0' (the character for the numeral for zero). In the latter case (as you suggest in the comment), you want string[i] < '0' || string[i] > '9'. (The numerals are guaranteed to be contiguous and ordered in any text encoding, so this works on any platform.)

Answer 4

You could use collection membership:

!boost::empty(boost::find<boost::return_found_end>("0123456789", string.at(i)))

Answer 5

For the C++ answer, look at this question which has already solved a quite similar problem, that you can very easily adapt to your situation.

As for Java, you can do this :

public boolean isInteger(String s) {
    return s.matches("^[0-9]+$");
}

You can modify the regex to suite your requirements. For example : "^[4-8]+$".

Note: String.matches is not optimal. If you need to perform checks often, use a compiled pattern instead :

static final Pattern DIGITS = Pattern.compile("^[0-9]+$");

public void isInteger(String s) {
    return DIGITS.matcher(s).find();
}

Answer 6

C++:

#include <cctype>
using namespace std;
...
if (!isdigit(str[i]))

// or

if (str[i] < '0' || str[i] > '9')

Java:

if (!Character.isDigit(str.charAt(i)))