C/C++ C++ - Determining If A String Contains Any Numeric Digits

[needOfHelpCMath]

Set hasDigit to true if the 3-character passCode contains a digit.

#include <iostream>
#include <string>
#include <cctype>
using namespace std;

int main() {
	bool hasDigit = false;
	string passCode;
	int valid = 0;

	passCode = "abc";

	if (hasDigit) {
		cout << "Has a digit." << endl;
	}
	else {
		cout << "Has no digit." << endl;
	}

	return 0;
}

Note: I cannot use loops.

 

[johng1]

Re: i am lost may anyone guide me or show me what to use

You need to be able to determine if a given char value is a digit. In header file cctype, function isdigit is precisely what you want. Here's one solution:

hasDigit = isdigit(passCode[0]) || isdigit(passCode[1]) || isdigit(passCode[2]);

If you don't yet know about the boolean operator || (or), think about using if statement(s).

 

[Greg]

Re: i am lost may anyone guide me or show me what to use

You could use

find_first_of

. There's an reference with an example here.

 

[needOfHelpCMath]

Re: i am lost may anyone guide me or show me what to use

You need to be able to determine if a given char value is a digit. In header file cctype, function isdigit is precisely what you want. Here's one solution:

hasDigit = isdigit(passCode[0]) || isdigit(passCode[1]) || isdigit(passCode[2]);

If you don't yet know about the boolean operator || (or), think about using if statement(s).

is it possible to use if statements to solve this program

 

[johng1]

First you need to find if passCode[0] is a digit:

hasDigit=false;
if (isdigit(passCode[0])) {
  hasDigit=true;
}

I hope you see that the above code is equivalent to:

hasDigit=isdigit(passCode[0]);

Next you need to test if passCode[1] is a digit:

if (isdigit(passCode[1])) {
  hasDigit=true;
}

So the following code tests whether passCode[0] or passCode[1] is a digit:

hasDigit=isdigit(passCode[0]);
if (isdigit(passCode[1])) {
  hasDigit=true;
}

I hope you see that the above code is not the same as:

hasDigit=isdigit(passCode[0]);
hasDigit=isdigit(passCode[1]);

Now you can finish with one more if statement.