How to extract year from IC number ?

Question

0.00/5 (No votes)

See more:

I have no idea how to extract year from IC number under convertYearofBirth and convert it from string to int.
Example IC : 190508143749 where '19' refers year, '05' refers month, '08' refers DOB.

What I have tried:

C++

#include <iomanip>
#include <string>
#include <cstring>
using namespace std;

bool checkCharacter(string);
int convertYearofBirth(string);
void checkEligibility(string, int);

int main()
{
	string name, IC;
	int cont = 1, year;

	do
	{
		cout << "Name: ";
		getline(cin, name);
		cout << "IC No (without hypen): ";
		cin >> IC;

		if (checkCharacter(IC))
		{
			year = convertYearofBirth(IC);

			checkEligibility(name, year);

		}
		else
			cout << "Please key in the correct IC no!" << endl;

		cout << "Cont (1-yes, 2-no)?";
		cin >> cont;
		cin.ignore();
		cout << fixed << setprecision(2) << endl;

	} while (cont == 1);

	return 0;
}

bool checkCharacter(string IC)
{
	for (int i = 0; i < IC.length(); i++)
		if (isdigit(IC[i]) == false)
			return false;
	return true;
}

int convertYearofBirth(string IC)
{
	for (int i = 0; i < 10; i++)
	{
		IC = stoi(IC.substr(0, 1));

	} 
}

void checkEligibility(string name, int year)
{
	if (year <= 2021)
	{
		cout << name << " is " << year << " years old." << endl;
		cout << name << " is eligible to vote! " << endl;
	}
	else
	{
		cout << name << " still has " << year << " to reach 21 years old." << endl;
		cout << name << " is not eligible to vote! " << endl;
	}
}

Posted 7-May-22 7:58am

Avatar 2 2

Updated 7-May-22 14:22pm

Patrice T

v2

Add a Solution

Comments

Richard MacCutchan 8-May-22 4:11am

If you know that the first six digits ar the year, month and day, then it is a simple matter to extract each pair of digits and convert them to integers.

2 solutions

Solution 2

The function to find a year is broken. It might look like this.

C++

int convertYearofBirth(const std::string &IC)
{
	int x;
	// x = stoi(IC.substr(0, 2));
	std::istringstream(IC.substr(0, 2)) >> x; 
	return x;
}

The other functions work exactly according to the same scheme.

Posted 7-May-22 14:22pm

merano99

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Rick York · Accepted Answer · 2022-05-07T10:45:00

Here is one simple and primitive way.

C++

void TestDayString()
{
	std::string IC( "190508143749" );

	char yearStr[ 4 ] = { 0 };
	char monthStr[ 4 ] = { 0 };
	char dayStr[ 4 ] = { 0 };

	strncpy( yearStr,  & IC[0], 2 );
	strncpy( monthStr, & IC[2], 2 );
	strncpy( dayStr,   & IC[4], 2 );

	int year = atoi( yearStr );
	int month = atoi( monthStr );
	int day = atoi( dayStr );

	printf( "year, month, days are %02d %02d %02d\n", year, month, day );
}

as an added bonus, I tried it and it works.