I used function overload to check if an input number is integer or float. However I get this following error:
error: call of overloaded 'retNr(double)' is ambiguous|
#include <iostream>
using namespace std;
void retNr(int x)
{
cout << "The entered number is an integer. " << endl;
}
void retNr(float x)
{
cout << "The entered number is a float. " << endl;
}
int main()
{
cout << "Please enter a number: " << endl;
cin >> nr;
retNr(nr);
return 0;
}
Read from cin into a string and then check the string for the presence of a decimal point. If there is a decimal point, call atof() on the string to convert it to a float, otherwise call atoi() to convert it to an integer.
Make some small change in:
void retNr(double x)
{
cout << "The entered number is a double. " << endl;
}
Remember to declare your nr variable.
double d = 1.0;
int i = 1;
retNr(d);
retNr(i);
You will have to initialize nr first.
Then you can use integer read & check it with a float if there is dot, ie ch=='.'
Thus, your program will be like this:
#include <iostream>
using namespace std;
int main()
{
int nr = 0; char ch;
cout << "Please enter a number: " << endl;
cin >> nr;
cin.get(ch);
if(ch=='.')
{
cout << "The entered number is a float. " << endl;
}
else
{
cout << "The entered number is an integer. " << endl;
}
return 0;
}
It's not too clear what you're asking for. If you really want
to know whether a number is an integer or not, then use modf
on it:
bool
isInt( double d )
{
double dummy;
return modf( d, &dummy ) == 0.0;
}
If you're reading a number, then read it as a double, and then
use the above.
If you want to trigger off the format of the input (i.e.
"10.0" will be treated as a floating point, even though it is
an integer), then read the input as a string, then try to
convert it to int; if this eats all of the input, then it was
entered as an int (no decimal or exponent), otherwise, try the
same thing treating it as a double:
std::string entry;
std::cin >> entry;
char const* end;
long i = strtol( entry.c_str(), &end, 10 );
if ( *end == '\0' ) {
// entry was entered in integral format...
} else {
double d = strtod( entry.c_str(), &end );
if ( *end == '\0' ) {
// entry was entered in floating point format...
} else {
// entry wasn't a number...
}
}
I'd advise against this, however; it will only confuse your
users if 0 isn't 0.0.
You may use abs() function for this issue.
#include<stdio.h>
#include<math.h>
int main()
{
double input;
scanf("%lf",&input);
int absulate = abs(input);
printf( (input==absulate)? "It is integer\n" : "It is float");
return 0;
}
float num = 7;
int n = (int)num;
float n1 = (float)n;
if(num == n1)
{
cout << "Integer\n";
}
else
{
cout << "Not Integer\n";
}
The question is wrong in its essence: A number it is not a float or an integer, but could be represented as a float or as an integer ( of course certain representation has some limitations )
So if I wrote '10' why should I say this is an integer? Could be a float too! Just if I want to use it as a float I would represent it as a float.
Related
I input a number in char type variable. like 12 or 22. but, console show me a 1 or 2.
How i get a whole number 12 ,22 in console?
#include <iostream>
int main()
{
using namespace std;
char a = 0;
cin >> a;
cout << a << endl;
return 0;
}
Here is console result.
12
1
C:\Users\kdwyh\source\repos\MyFirstProject\Debug\MyFirstProject.exe(프로세스 18464개)이(가) 종료되었습니다(코드: 0개).
이 창을 닫으려면 아무 키나 누르세요...
The reason I don't use int, string and something is because I want to get both number and Character in one variable.
So I want to see the results of combined numbers and character at the same time.
in that process i can't get a whole number.
#include <iostream>
using namespace std;
int index = 0;
constexpr int pagenum = 10;
void chapterlist(void);
void nextlist(void);
void beforelist(void);
void movechapter(char a);
int main(void)
{
char userin = 0;
bool toggle = 0;
cout << "결과를 볼 챕터를 고르시오." << endl;
chapterlist();
cout << "다음 페이지로 이동: n" << endl;
cin >> userin;
if (userin == 'n')
{
backflash:
while(toggle == 0)
{
nextlist();
cin >> userin;
if (userin == 'b')
{
toggle = 1;
goto backflash;
}
else if (userin == 'n')
continue;
else
{
system("cls");
movechapter(userin);
break;
}
}
while(toggle == 1)
{
beforelist();
cin >> userin;
if (userin == 'n')
{
toggle = 0;
goto backflash;
}
else if (userin == 'b')
continue;
else
{
system("cls");
movechapter(userin);
break;
}
}
}
else
{
system("cls");
movechapter(userin);
}
return 0;
}
void chapterlist(void)
{
int x = 0;
for (x = index + 1; x <= index + 10; x++)
cout << "Chapter." << x << endl;
}
void nextlist(void)
{
system("cls");
cout << "결과를 볼 챕터를 고르시오." << endl;
index = index + pagenum;
chapterlist();
cout << "다음 페이지로 이동: n" << endl;
cout << "이전 페이지로 이동: b" << endl;
}
void beforelist(void)
{
system("cls");
cout << "결과를 볼 챕터를 고르시오." << endl;
index = index - pagenum;
chapterlist();
cout << "다음 페이지로 이동: n" << endl;
cout << "이전 페이지로 이동: b" << endl;
}
void movechapter(char a)
{
cout << "선택한 Chapter." << a << "의 결과입니다." << endl;
}
In movechapter(), console show me a is 1 or 2, not 12, 22.
First, you have to understand what achar type is.
Character types: They can represent a single character, such as 'A' or '$'. The most basic type is char, which is a one-byte character. Other types are also provided for wider characters.
To simplify that, char can only hold one character.
Where as with your code, "12" is actually 2 separate characters, '1' and '2', and that's the reason it would not work.
Instead of declaring a as a char type, you could declare it as an int type, which is a type designed to hold numbers. So you would have:
int a = 0;
However, do note that int often has a maximum value of 2^31.
Or you could use std::string to store character strings. However, do note that if you wish to do any calculations to your string type, you would need to convert them to a number type first:
int myInt = std::stoi(myString);
Edit:
So I have re-checked your code after your update, there is nothing wrong with using std::string in your case. You can still check if user have input n or b by:
if (userin == "n")
Note that you would use double quotation mark, or "letter", around the content that you want to check.
On the other hand, you could use:
if(std::all_of(userin .begin(), userin.end(), ::isdigit))
To check if user have input a number.
Although char is just a number, it's presumed to mean "single character" here for input. Fix this by asking for something else:
int a = 0;
You can always cast that to char as necessary, testing, of course, for overflow.
You should be reading characters into a string, and then converting that string into an int. It would also probably make more sense to use something like getline() to read input, rather than cin >> a.
#include <string>
#include <iostream>
#include <stdexcept>
#include <stdio.h>
int main() {
std::string input_string;
/* note that there is no function that will convert an int string
to a char, only to an int. You can cast this to a char if needed,
or bounds check like I do */
int value;
while(1) {
getline(std::cin, input_string);
/* std::stoi throws std::invalid_argument when given a string
that doesn't start with a number */
try {
value = std::stoi(input_string);
} catch (std::invalid_argument) {
printf("Invalid number!\n");
continue;
}
/* You wanted a char, the max value of a `char` is 255. If
you are happy for any value, this check can be removed */
if (value > 255) {
printf("Too big, input a number between 0-255\n");
continue;
}
break;
}
printf("Number is %hhu\n", value);
}
I wrote code to convert decimal fraction number to its binary equivalent. It compiles fine but when executed hangs. The code here prints only first four digits of the binary conversion and if the number if with more than 4 digits, it shows '...' after it. On execution it hangs. Help!
#include <iostream>
using namespace std;
int main()
{
int i, x[10];
float num;
cout << "**PROGRAM TO CONVERT DECIMAL FRACTION INTO ITS EQUIVALENT BINARY**\n";
cout << "Enter a fraction in between 0 to 1 for conversion: ";
cin >> num;
if (num>=0 && num<=1)
{
i=1;
while (num!=1.000)
{
num*=2;
x[i]=num;
num=num-x[i];
i++;
}
if (i>4)
{
cout << "The binary equivalent is 0.";
for (i=1;i<=4;i++)
{
cout << x[i];
}
cout << "...";
}
else
{
cout << "The binary equivalent is 0.";
for (i=1;i<=4;i++)
{
cout << x[i];
}
}
}
else
{
cout << "The number entered is out of range.";
}
return 0;
}
The first obstacle is the infinite while loop:
Assuming input num=0.5
after first iteration, i=1, x[0]=1, num=0.0
after second iteration, i=2, x[1]=0, num=0.0
Continue forever, i=..., x[i-]1=0, num=0.0
With nothing to break the loop.
while (num!=1.000)
{
num*=2;
x[i]=num;
num=num-x[i];
i++;
}
To fix, consider few changes. There might be other issues.
put a limit on the while loop (i<10 should be a good condition, as this is the size of the x array), or i=4, as this is the maximum output.
The break condition for the while loop should probably be 'num != 0', or even better (num > 1e-7, or other small value).
float has 23 bit in mantissa, maybe it is because you are assign x[i] with i greater than 9.
try this:
//stop when you get four bits
while (i< 5)
Original code has several issues:
1 For input num=.5 and similar (really for all values) cycle never ends (dash-o suggested fix ideas)
2 array x[10] is overflowed with undefined behavior (Edney)
3 nitpicking: 1 is not a “fraction” and better check for a range 0 <= num < 1 instead of 0 <= num <= 1(see also OP printing code; 1 could be added); we could use x[4] with 0<=i <=3
4 string could also be used (PaulMcKenzie). Really “>>” uses string processing for parsing and calculating binary equivalent from which by multiplying by 2 (left shit) and truncation fractional part the code calculates target bits. Both approaches give correct identical results; implementing by string we need to add internal to operator “>>” implementation code to parsing valid formats for floats (decimals) such as 3.14e-1, .2718, 1e-1, etc.
This fix follows OP:
#include <iostream>
using namespace std;
int main()
{
int i, x[5];
float num;
cout << "**PROGRAM TO CONVERT DECIMAL FRACTION INTO ITS EQUIVALENT BINARY**\n";
cout << "Enter a fraction in between 0 to 1 for conversion: ";
cin >> num;
if (num>=0 && num<1)
{
i=1;
while (i<=4)
{
num*=2;
x[i]=num;
num=num-x[i];
i++;
}
cout << "The binary equivalent is 0.";
for (i=1;i<=4;i++)
{
cout << x[i];
}
if (num>0)
cout << "...";
}
else
{
cout << "The number entered is out of range.";
}
return 0;
}
This code is without cycles (they are in code implementing “>>”, bitset):
#include <iostream>
#include <bitset>
using namespace std;
int main () {
const int digits = 4;
int fraction;
float num;
cout << "**PROGRAM TO CONVERT DECIMAL FRACTION INTO ITS EQUIVALENT BINARY**\n";
cout << "Enter a fraction in between 0 to 1 for conversion: ";
cin >> num;
if (num >= 0 && num < 1) {
fraction = num = num * pow (2, digits);
cout << "The binary equivalent is 0.";
cout << bitset<digits> (fraction);
if (num - fraction > 0) cout << "...";
}
else cout << "The number entered is out of range.";
}
I was typing this and it asks the user to input two integers which will then become variables. From there it will carry out simple operations.
How do I get the computer to check if what is entered is an integer or not? And if not, ask the user to type an integer in. For example: if someone inputs "a" instead of 2, then it will tell them to reenter a number.
Thanks
#include <iostream>
using namespace std;
int main ()
{
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}
You can check like this:
int x;
cin >> x;
if (cin.fail()) {
//Not an int.
}
Furthermore, you can continue to get input until you get an int via:
#include <iostream>
int main() {
int x;
std::cin >> x;
while(std::cin.fail()) {
std::cout << "Error" << std::endl;
std::cin.clear();
std::cin.ignore(256,'\n');
std::cin >> x;
}
std::cout << x << std::endl;
return 0;
}
EDIT: To address the comment below regarding input like 10abc, one could modify the loop to accept a string as an input. Then check the string for any character not a number and handle that situation accordingly. One needs not clear/ignore the input stream in that situation. Verifying the string is just numbers, convert the string back to an integer. I mean, this was just off the cuff. There might be a better way. This won't work if you're accepting floats/doubles (would have to add '.' in the search string).
#include <iostream>
#include <string>
int main() {
std::string theInput;
int inputAsInt;
std::getline(std::cin, theInput);
while(std::cin.fail() || std::cin.eof() || theInput.find_first_not_of("0123456789") != std::string::npos) {
std::cout << "Error" << std::endl;
if( theInput.find_first_not_of("0123456789") == std::string::npos) {
std::cin.clear();
std::cin.ignore(256,'\n');
}
std::getline(std::cin, theInput);
}
std::string::size_type st;
inputAsInt = std::stoi(theInput,&st);
std::cout << inputAsInt << std::endl;
return 0;
}
Heh, this is an old question that could use a better answer.
User input should be obtained as a string and then attempt-converted to the data type you desire. Conveniently, this also allows you to answer questions like “what type of data is my input?”
Here is a function I use a lot. Other options exist, such as in Boost, but the basic premise is the same: attempt to perform the string→type conversion and observe the success or failure:
template <typename T>
auto string_to( const std::string & s )
{
T value;
std::istringstream ss( s );
return ((ss >> value) and (ss >> std::ws).eof()) // attempt the conversion
? value // success
: std::optional<T> { }; // failure
}
Using the optional type is just one way. You could also throw an exception or return a default value on failure. Whatever works for your situation.
Here is an example of using it:
int n;
std::cout << "n? ";
{
std::string s;
getline( std::cin, s );
auto x = string_to <int> ( s );
if (!x) return complain();
n = *x;
}
std::cout << "Multiply that by seven to get " << (7 * n) << ".\n";
limitations and type identification
In order for this to work, of course, there must exist a method to unambiguously extract your data type from a stream. This is the natural order of things in C++ — that is, business as usual. So no surprises here.
The next caveat is that some types subsume others. For example, if you are trying to distinguish between int and double, check for int first, since anything that converts to an int is also a double.
There is a function in c called isdigit(). That will suit you just fine. Example:
int var1 = 'h';
int var2 = '2';
if( isdigit(var1) )
{
printf("var1 = |%c| is a digit\n", var1 );
}
else
{
printf("var1 = |%c| is not a digit\n", var1 );
}
if( isdigit(var2) )
{
printf("var2 = |%c| is a digit\n", var2 );
}
else
{
printf("var2 = |%c| is not a digit\n", var2 );
}
From here
If istream fails to insert, it will set the fail bit.
int i = 0;
std::cin >> i; // type a and press enter
if (std::cin.fail())
{
std::cout << "I failed, try again ..." << std::endl
std::cin.clear(); // reset the failed state
}
You can set this up in a do-while loop to get the correct type (int in this case) propertly inserted.
For more information: http://augustcouncil.com/~tgibson/tutorial/iotips.html#directly
You can use the variables name itself to check if a value is an integer.
for example:
#include <iostream>
using namespace std;
int main (){
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
if(firstvariable && secondvariable){
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}else{
cout << "\n[ERROR\tINVALID INPUT]\n";
return 1;
}
return 0;
}
I prefer to use <limits> to check for an int until it is passed.
#include <iostream>
#include <limits> //std::numeric_limits
using std::cout, std::endl, std::cin;
int main() {
int num;
while(!(cin >> num)){ //check the Input format for integer the right way
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
cout << "Invalid input. Reenter the number: ";
};
cout << "output= " << num << endl;
return 0;
}
Under C++11 and later, I have the found the std::stoi function very useful for this task. stoi throws an invalid_argument exception if conversion cannot be performed. This can be caught and handled as shown in the demo function 'getIntegerValue' below.
The stoi function has a second parameter 'idx' that indicates the position of the first character in the string after the number. We can use the value in idx to check against the string length and ascertain if there are any characters in the input other than the number. This helps eliminate input like 10abc or a decimal value.
The only case where this approach fails is when there is trailing white space after the number in the input, that is, the user enters a lot of spaces after inputting the number. To handle such a case, you could rtrim the input string as described in this post.
#include <iostream>
#include <string>
bool getIntegerValue(int &value);
int main(){
int value{};
bool valid{};
while(!valid){
std::cout << "Enter integer value: ";
valid = getIntegerValue(value);
if (!valid)
std::cout << "Invalid integer value! Please try again.\n" << std::endl;
}
std::cout << "You entered: " << value << std::endl;
return 0;
}
// Returns true if integer is read from standard input
bool getIntegerValue(int &value){
bool isInputValid{};
int valueFromString{};
size_t index{};
std::string userInput;
std::getline(std::cin, userInput);
try {
//stoi throws an invalid_argument exception if userInput cannot be
//converted to an integer.
valueFromString = std::stoi(userInput, &index);
//index being different than length of string implies additional
//characters in input that could not be converted to an integer.
//This is to handle inputs like 10str or decimal values like 10.12
if(index == userInput.length()) isInputValid = true;
}
catch (const std::invalid_argument &arg) {
; //you could show an invalid argument message here.
}
if (isInputValid) value = valueFromString;
return isInputValid;
}
You could use :
int a = 12;
if (a>0 || a<0){
cout << "Your text"<<endl;
}
I'm pretty sure it works.
I am a newbie to C++. I have a situation where the input integer is taken from the user. However, I need to check if the user enters a decimal value. How do I check this?
I have tried cin.good(), cin.fail() but they are detecting only non-digit entries and not decimal numbers. Any help would be appreciated.
#include <iostream>
int main()
{
using namespace std;
int x;
cout << "Enter an integer: " << endl;
cin >> x;
if (cin.good()) {
cout << "input is an integer" << endl;
}
else
cout << "input is not an integer" << endl;
}
Here's my output:
1.
Enter an integer:
1.2
input is an integer
2.
Enter an integer:
a
input is not an integer
float x = 4.2;
if (x == (int) x)
{
// int
}
else
{
// not int
}
You can use std::isdigit for checking your string input next way.
bool is_numeric(const std::string& str)
{
std::string::const_iterator it = str.begin();
if (it != str.end() && *it == '-') ++it;
if (it == str.end()) return false;
while (it != str.end() && std::isdigit(*it)) ++it;
return it == str.end();
}
It's not hard to change it to work with floating points, if needs, but that function will exactly checks what you need.
You receive the input as an int from cin and hence any float entered would already be truncated by the time you get your hands on it. You should receive it as a float or a string to decide on the validity of the input.
Removed the earlier answer since it went down the slippery route of manually parsing the input which is unnecessary and error-prone. The standard library already has multiple ways to check if an input is a valid number. Two ways that I know: C++ streams and the C library function strtof. Here's an example using the latter:
#include <iostream>
#include <string>
#include <cmath>
bool is_int(float f) {
return std::floor(f) == f;
}
int main()
{
std::cout << "Enter an integer: ";
std::string input;
std::cin >> input;
char *e = nullptr;
char const *str = input.c_str();
float const f = strtof(str, &e);
// no conversion was performed or was stopped as disallowed
// characters were encountered: Not A Number
if ((e == str) || (*e != '\0'))
std::cout << "NAN";
else if ((f == HUGE_VALF) || !std::isfinite(f))
std::cout << "too large";
else
std::cout << (is_int(f) ? "integer" : "non-integer");
std::cout << '\n';
}
Live example.
To check if the input is a number, this
float f;
cin >> f;
is possible too, but it will also accept NANs as valid input e.g. 45dsf will be converted to 45. One has to then check if the conversion happened completely and successfully by checking the fail and eof bits of the stream.
See also
Checking if float is an integer
C++ IsFloat function
How to convert a number to string and vice versa in C++
I've searched for this answer, and no one seems to know how to fix this error. I want the input to be strictly an int. If the input is a double, I want it to send an error.
int creatLegs = 0;
string trash;
bool validLegs = true;
do
{
cout << "How many legs should the creature have? ";
cin >> creatLegs;
if(cin.fail())
{
cin.clear();
cin >> trash; //sets to string, so that cin.ignore() ignores the whole string.
cin.ignore(); //only ignores one character
validLegs = false;
}
if (creatLegs > 0)
{
validLegs = true;
}
if (!validLegs)
{
cout << "Invalid value, try again.\n";
}
} while (!validLegs);
It seems to almost work. It sends the error, but only after moving onto the next loop. How can I fix this? And why is it still showing the error message but still moving on before showing it?
An input can be something else than a representation of an integer or of a floating point number.
Remember that numbers are not their representation(s): 9 (decimal), 017 (octal, à la C), 0b1001 (binary, à la Ocaml), IX (Roman notation), 8+1 (arithmetic expression), neuf (French) are all representations of the same number nine.
So you have to decide if you accept an input like 9 x, or 9 (with several spaces after the digit), ... More generally you have to define what are the acceptable inputs (and if the input is ending at end of line or not, if spaces or punctuation should be accepted, etc...).
You could read an entire line (e.g. with std::getline) and use e.g. sscanf (where the %n control format is useful, and so is the item count returned by sscanf) or std::stol (where you use the end pointer) to parse it
Notice also that the phrasing of your question ("Distinguishing between an int and a double") is wrong. There is no single "int or double" type in C++ (but int is a scalar type, and double is a scalar type in C++, and you could define a class with a tagged union to hold either of them). AFAIU, if you declare int x; then use std::cin >> x; with the user inputting 12.64 the dot and the digits 64 after it won't be parsed and x would become 12.
I think that you should read data as string, and then check it char by char to verify that it is integer - if every char is a digit, then we have integer and we can parse it.
Problem with streams is, that if you're trying to read integer but decimal is passed, it reads the number up to the dot. And this part is a proper integer, so cin.fail() returns false.
Sample code:
#include <iostream>
#include <string>
#include <cctype>
#include <cstdlib>
using namespace std;
int main() {
int creatLegs = 0;
bool validLegs = true;
do
{
cout << "How many legs should the creature have? ";
string input;
getline(cin, input);
validLegs = true;
for (string::const_iterator i = input.begin(); validLegs && i != input.end(); ++i) {
if (!isdigit(*i)) {
validLegs = false;
}
}
if (!validLegs)
{
cout << "Invalid value, try again.\n";
} else {
creatLegs = atoi(input.c_str());
}
} while (!validLegs);
cout << creatLegs << endl;
}
This of course is not a perfect solution. If there any leading or trailing spaces (or any other characters like + or -), the program will fail. But you always can add some code to handle those situations, if you need to.
int creatLegs = 0;
do
{
cout << "How many legs should the creature have? ";
cin >> creatLegs; // trying to get integer
if(!cin.fail()) // if cin.fail == false, then we got an int and leave loop
break;
cout << "Invalid value, try again.\n"; // else show err msg and try once more
cin.clear();
} while (1);
This question already has an accepted answer, however I'll contribute a solution that handles all numbers that are integral, even those that are expressed as a floating point number (with no fractional part) and rejects input that contains anything other than spaces following the number.
Examples of accepted values, these all represent the number 4:
4
4.
4.0
+4
004.0
400e-2
Examples of rejected values:
3.999999
4.000001
40e-1x
4,
#include <iostream>
#include <sstream>
#include <cctype>
#include <string>
using namespace std;
bool get_int( const string & input, int & i ) {
stringstream ss(input);
double d;
bool isValid = ss >> d;
if (isValid) {
char c;
while( isValid && ss >> c ) isValid = isspace(c);
if (isValid) {
i = static_cast<int>(d);
isValid = (d == static_cast<double>(i));
}
}
return isValid;
}
int main( int argc, char *argv[] )
{
int creatLegs = 0;
bool validLegs = false;
do
{
string line;
do {
cout << "How many legs should the creature have? ";
} while (not getline (cin,line));
validLegs = get_int( line, creatLegs );
if (creatLegs <= 0)
{
validLegs = false;
}
if (not validLegs)
{
cout << "Invalid value, try again." << endl;
}
} while (not validLegs);
cout << "Got legs! (" << creatLegs << ")" << endl;
return 0;
}
If you want strictly integers (no decimal period and no scientific notation) then use this simpler get_int function:
bool get_int( const string & input, int & i ) {
stringstream ss(input);
bool isValid = ss >> i;
if (isValid) {
char c;
while(isValid && ss >> c) isValid = isspace(c);
}
return isValid;
}