Check if a string is a number in C++

In this article, we will discuss different ways to check if a string is a number or not in C++. Basically, we will confirm that string contains only digits.

Table Of Contents

Problem description

In this article, we have given a string and we have to check whether the string is a number or not with the help of C++ STL. Our C++ program can iterate the string and check if all the characters of string is a number or not by different methods.

Input

string s1 = "54321";
string s2 = "-12345";
string s3 = "a1b2c3";

Output

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s1 is a Number
s2 is a Number
s3 is not a Number

Method 1: Using Loop

In this method we are using a loop to check whether every character is a number or not. There are three string declared in the program and each string is passed in isNumeric() function which is a boolean function and return either true or false. Inside the isNumeric() function we are iterating the string with the help of iterator. Increment of iterator is done only if current character of string is a number, which can be checked by isdigit() function. The isNumeric function returns true only if the iterator reached to end of the string. This method fails to give correct answer in case of negative number present in string.

Time Complexity: O(n)
Space Complexity: O(1)

#include <iostream>
#include <string>

using namespace std;

bool isNumber(string str)
{
    // Declare the string iterator
    string::iterator it = str.begin();

    // Iterating in string with loop
    while (it != str.end() && isdigit(*it))
    {
        // Increment the iterator
        it++;
    }
    // Return true if iterator goes to end of string
    return !str.empty() && it == str.end();
}

int main()
{
    // Define three strings
    string s1 = "54321";
    string s2 = "-12345";
    string s3 = "a1b2c3";

    // check the string is a number by calling the isNumeric() function
    cout << "s1 is " << (isNumber(s1) ? "a Number" : "not a Number") << endl;
    cout << "s2 is " << (isNumber(s2) ? "a Number" : "not a Number") << endl;
    cout << "s3 is " << (isNumber(s3) ? "a Number" : "not a Number") << endl;

    return 0;
}

Output

s1 is a Number
s2 is not a Number
s3 is not a Number

Method 2: Using STL Algorithm std::find_if()

In this method we are using find_if() and check whether every character is a number or not. There are three string declared in the program and each string is passed in isNumeric() function which is a boolean function and return either true or false. Inside the isNumeric() function we are iterating the string with find_if() function. The find_if() function have three arguments: begin iterator, end iterator and boolean function. In the boolean function the characters of string are checked whether they are numbers or not with the help of isdigit() function. The find_if() function return the iterator which points the character of string where the boolean function (of find_if()) returns flase. The isNumeric() function returns true only if the iterator reached to end of the string. This method fails to give correct answer in case of negative number present in string.

Time Complexity: O(n)
Space Complexity: O(1)

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

bool isNumber(string str)
{
    // Use of find_if() function
    auto it = find_if(str.begin(), str.end(), [](char const &c) {
        return !isdigit(c);
    });

    // Return true if iterator goes to end
    return !str.empty() && it == str.end();
}

int main()
{
    // Define three strings
    string s1 = "54321";
    string s2 = "-12345";
    string s3 = "a1b2c3";

    // check the string is a number by calling the isNumeric() function
    cout << "s1 is " << (isNumber(s1) ? "a Number" : "not a Number") << endl;
    cout << "s2 is " << (isNumber(s2) ? "a Number" : "not a Number") << endl;
    cout << "s3 is " << (isNumber(s3) ? "a Number" : "not a Number") << endl;

    return 0;
}

Output

s1 is a Number
s2 is not a Number
s3 is not a Number

Method 3: Using STL Algorithm std::all_of()

In this method we are using all_of() function and check whether every character is a number or not. There are three string declared in the program and each string is passed in isNumeric() function which is a boolean function and return either true or false. Inside the isNumeric() function we are iterating the string with the help of all_of() function which have 3 arguments: begin iterator, end iterator and boolean function. The boolean function used here is isdigit() function which checks whether a character is numeric or not. This way all_of() function returns true only if all the characters in string is numeric. The isNumeric() function returns true only if the all_of() function returns true. This method fails to give correct answer in case of negative number present in string.

Time Complexity: O(n)
Space Complexity: O(1)

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

bool isNumber(string str)
{
    // Return true if all elemets of string is a number
    return !str.empty() && all_of(str.begin(), str.end(), ::isdigit);
}

int main()
{
    // Define three strings
    string s1 = "54321";
    string s2 = "-12345";
    string s3 = "a1b2c3";

    // check the string is a number by calling the isNumeric() function
    cout << "s1 is " << (isNumber(s1) ? "a Number" : "not a Number") << endl;
    cout << "s2 is " << (isNumber(s2) ? "a Number" : "not a Number") << endl;
    cout << "s3 is " << (isNumber(s3) ? "a Number" : "not a Number") << endl;

    return 0;
}

Output

s1 is a Number
s2 is not a Number
s3 is not a Number

Method 4: Using strtol() function

In this method we are using strtol() function and check whether every character is a number or not. There are three string declared in the program and each string is passed in isNumeric() function which is a boolean function and return either true or false. Inside the isNumeric() function we are using strtol() function which is used to convert the string into a number with a specified base. The strtol function takes 3 arguments as a input: C-string, char pointer and base of number (for decimal number base is 10). The conversion of string to C-string is done by c_str() function.
The isNumeric() function returns true only if character pointer is null. This approach is good for positive as well as negative numbers present in the string.

Time Complexity: O(n)
Space Complexity: O(1)

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

bool isNumber(string str)
{
    // Declare the char pointer
    char* p;

    // Use of strtol() function with C-string, char pointer and base as a input
    strtol(str.c_str(), &p, 10);

    // Return true if char pointer p is null
    return *p == 0;
}

int main()
{
    // Define three strings
    string s1 = "54321";
    string s2 = "-12345";
    string s3 = "a1b2c3";

    // check the string is a number by calling the isNumeric() function
    cout << "s1 is " << (isNumber(s1) ? "a Number" : "not a Number") << endl;
    cout << "s2 is " << (isNumber(s2) ? "a Number" : "not a Number") << endl;
    cout << "s3 is " <<  (isNumber(s3) ? "a Number" : "not a Number") << endl;

    return 0;
}

Output

s1 is a Number
s2 is a Number
s3 is not a Number

Summary

Today, we learned about various methods to check if a string is a number or not. Each method has its own time and space complexity. Thanks.

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Thanks for reading.

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