Implement isNumber()

The code is similar to atop but here we also need to check if number is a double. We need to make sure that the string has at most one decimal point.
The key steps in writing an isNumber function are:

• Check for sign(-) after we've checked for whitespace
• Check if each character lies between 0 & 9
• Make note of first decimal encountered.
• Return false if we encounter a second decimal point
• Return false as soon as we see a non numeric character
• Return true otherwise.

bool isNumber(string toTest)
{
    int index =0, decimal = 0;

    //Checking for sign
    if(toTest[index] == '-')
        index++;
    
    while(index<toTest.length())
    {
        int num = toTest[index] - '0';

        //Check for numerical digit
        if(num >=0 && num <=9)
        {
            index++;
            continue;
        }

        //Check if character is a decimal point and if its the first decimal point.
        else if(toTest[index] == '.')
        {

            //Number cannot have 2 decimal points so return false.
            if(decimal)
                return false;
            else
                decimal = 1;
            index++;
        }
        else
        return false;
    }
    return true;
}

Implement atoi()

The key steps in writing an atoi function are:

• Check for whitespace at the start of the string
• Check for sign(-) after we've checked for whitespace
• Check if each character lies between 0 & 9
• Break as soon as we see a non numeric character
• Make sure to multiply by sign before returning result

int atoi(const char *text)
{
    if(text==NULL || strlen(text)==0)
    return 0;
    
    int len = strlen(text), i = 0, sign = 1, num = 0;
    
    //Checking for whitespace
    while(text[i] == ' ' || text[i] == '\t' || text[i] == '\n')
    i++;
    
    //Checking for sign
    if(text[i] == '-')
    {
        sign=-1;
        i++;
    }
    
    //Checking for numeric characters
    while(i<len)
    {
        if((text[i] - '0') >=0 && (text[i] - '0') <=9)
        {
            num = num*10 + (text[i] -'0');
            i++;
        }
        //Stopping when we see non numeric character
        else
        break;
    }
    
    // multiply the number by sign before returning
    return num*sign;
}