c function/code to test number

[babble]

So I am reading in a number from a string.

I need to check if that number is valid.

Invalid examples: "3e343" - "ww3" - "2.2.2" - "33 44" - "fafa" - "+" - "-" - "324+1"
Valid examples: "343" - "-3242" - "+3242"

Thanks!

[Dheeraj Joshi]

Take the number as string. check for each character whether it is a number or character or allowed symbol or white space. Do the operations as required.

Regards
Dheeraj Joshi

[donbock]

Standard function strtol will do most of the work for you. It is up to you to use the endptr argument to verify there are no trailing nondigit characters.

[babble]

Can you be more detailed in your response?

[donbock]

You need to make a stab at implementing this first. If you get stuck then come back here and describe your specific problem and we'll try to help.

Take a look at your first post. You didn't actually ask a question.

[babble]

I used if statements to check for different conditions, but, my if statement is getting way too large.

[donbock]

Take a look at the character-classification functions in Standard header ctype.h.

Your if statement can test for each character category, then call specific per-category functions as needed.

[whodgson]

What is being suggested above is that you use one of the available functions to convert the string to a numeral type such as an int. For example donbrock mentions strtol() for converting a string to a long type numeral. Another is atoi(const *char str) for converting a string to an integer. If str =" 21x.3",atoi(str) produces the integer 21. Whether it atoi() or strtol() will strip out or condense all the invalid string components leaving a 'valid' series of digits is something you will have to test for.
b.t.w. I do not understand what you are trying to achieve with all the minus operators.

[whodgson]

What is being suggested above is that you use one of the available functions to convert the string to a numeral type such as an int. For example donbrock mentions strtol() for converting a string to a long type numeral. Another is atoi(const *char str) for converting a string to an integer. If str =" 21x.3",atoi(str) produces the integer 21. Whether it atoi() or strtol() will strip out or condense all the invalid string components leaving a 'valid' series of digits is something you will have to test for.
b.t.w. I do not understand what you are trying to achieve with all the minus operators.

[jkmyoung]

I'm not sure if you've considered this, so forgive me if this sounds stupid:

if a character is >= '0' but <= '9' then it is a digit.

Would that help reduce your if else clauses?

[donbock]

@jkmyoung
This is only true if the character codes for the digits are consecutive and if '0' has the smallest value and '9' has the largest. This is true for ASCII, but the Standard does not require it to be true of all legal character encodings.

Assuming ASCII encoding is a good bet, but why gamble at all? The standard isdigit function accomplishes the same thing and guarantees portability.

[babble]

Basically, if the user enters any of the following strings:

"3e343" - "ww3" - "2.2.2" - "33 44" - "fafa" - "+" - "-" - "324+1"

They would get, "Invalid entry, please enter numerical values only"

On the other hand, if the user enters any of the following:

"343" - "-3242" - "+3242" - "4.5" - "330.6"

They would get - "Values accepted"

[emibt08]

bool isValidNumber(const std::string& str)
{
    size_t len = str.length();
 
    if (!len)
        return false;
 
    char ch = str[0];
 
    if (1 == len)
        return isdigit(ch);
 
    bool hasDot = false;
 
    if (!isdigit(ch) && '+' != ch && '-' != ch)
    {
        if ('.' == ch)
            hasDot = true;
        else
            return false;
    }
 
    bool hasDigit = false;
 
    while (--len)
    {
        ch = str[len];
 
        if ('.' == ch)
        {
            if (hasDot)
                return false;
            else
                hasDot = true;
        }
        else if (isdigit(ch))
            hasDigit = true;
        else
            return false;
    }
 
    return hasDigit;
}
 

[donbock]

emibt08, there are a couple of problems with your code:

• The point of this forum is to help the OP learn how to use C. Handing him a cut-and-paste solution to his homework problem rarely helps us reach that goal.
• Your program considers "+." and -." to be valid numbers.
• Your program considers ".5" to be invalid.

It helps with problems like this to draw a state diagram. This allows you to focus on exactly what needs to happen without worrying about how to implement it.

[emibt08]

@donbock
donbock, thanks for ponting that out.
I wanted to give to the user a simple function that does what he wants to accomplish since it's pretty simple. However, I guess I shouldn't have because of what you said about the forum's goal. I apologize for that.
since I already posted the function, I just fixed it to address the issues that you found in it.