How to check using isdigit() if there more than one character - c++

Question

So, I'm still new on coding, and my friend is kindly taught me a lot and I want to be learn it quick, then he gave me this exercise to study in my home
He gave me this question about booking table number, so I tried to make it..
But I'm having trouble with this part checking the table number if it is a digit or not... I was searching on google on how, but they only give example on how to do it with single character.. So, I made an account then ask here xD
btw, the table max number is 25.
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<ctype.h>
bool checkDigit(char tablenum){
char digit[3];
int n;
int value = tablenum;
if(tablenum>0 && tablenum<10)
n=1;
if(tablenum>9 && tablenum < 26)
n=2;
for(int i=0;i<n;i++){
digit[i] = value%10;
value /= 10;
}
if(n==1){
printf("digit 0 = %d\nisdigit = %d\n",tablenum,isdigit(tablenum)); //just checking the isdigit value
if(isdigit(digit[0])!=0){
puts("It's digit");
return true;
}
else{
puts("It's not digit");
return false;
}
}
if(n==2){
printf("digit 0 = %d isdigit 0 = %d \ndigit 1 = %d isdigit 1 = %d\n",digit[0],isdigit(digit[0]),digit[1],isdigit(digit[1])); //checking the values too
if(isdigit(digit[0]) != 0 && isdigit(digit[1]) != 0){
puts("It's digits!");
return true;
}
else{
puts("It's not digits!");
return false;
}
}
else{
printf("It's not digit!")
return false;
}
}
int main(){
int num;
bool itsdigit;
do{
scanf("%d",&num);fflush(stdin);
itsdigit = checkDigit(num);
} while(itsdigit != 1 );
getchar();
return 0;
}
Input = "25" "1" "0" "test"
Output = "It's a digit" "It's a digit" "It's not a digit" "
Since the table is only 1 to 25
When I input "1" on it shows the right digit, but when it when in to isdigit() I keep getting 0.
I wonder what when wrong with this, and please give explanation.
And thanks before!

Answer 1

You have multiple problems.
Lets start by taking a closer look at these two lines:
checkDigit(num);
itsdigit = checkDigit;
The first line calls checkDigit but throws away the result.
The second line does not call the function, instead the compiler will let it decay to a pointer to the function and assign that pointer to itsdigit. Your compiler should have complained about it.
Furthermore, isdigit doesn't have to return 1 when the character is a digit, it only have to return something which is non-zero.
Lastly, in the digits array you don't store characters, you store integer values. isdigit(1) will be false while isdigit('1') will be true. Just because you use the type char doesn't automatically store characters in the variables. And considering that you read an integer, and store its digits in the digits array, there's no need to check if the digits are actually digits. Of course the digits will be digits.

Answer 2

You probably want this:
bool IsOnlyDigits(const char text[])
{
int length = strlen(text);
for (int i = 0; i < length; i++)
{
if (!isdigit(text[i]))
return false;
}
return true;
}
int main() {
char num[50];
scanf("%s", num);
if (IsOnlyDigits(num))
puts("It's digits!");
else
puts("It's not digits!");
fflush(stdout);
getchar();
return 0;
}
Disclaimer: non error checking and non bounds checking code. #includes omitted for brevity. Is has been tested under Visual Studio 2015 and seems to be working fine here.
Examples of input and output:
123
It's digits!
12a
It's not digits!
abc
It's not digits!
To understand this, you need to learn about strings and arrays. Just read the corrsponding chapters in your C textbook.

Answer 3

You could instead create an array of type char or int, iterate through the array and check if any character instance is a digit or not using the isdigit() function.
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
char arr[10];
cin>>arr;
for(int i=0;i<strlen(arr);i++){
if(isdigit(arr[i])){
cout<<arr[i]<<" is a digit\n";
}
else {
cout<<arr[i]<<" is not a digit\n";
}
}
return 0;
}