Understanding of while loop and the difference of "! " and "false" in C++

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I'm trying to understand while loop, and the difference when I put "!" and "false" as a condition because somehow they produced different result for me.

I'm writing this very simple program to sum an array of numbers, and I want to check if my input is integer or not.

Flowchart :

1. Ask user how many numbers to sum

2. Get user input (check if this is integer or not)

3. (loop) ask user to input each value of number

4. sum them

What I have tried:

This is my code, I'm using `isdigit` to check whether an input is an integer, and check it using while (isdigit(array_size==false))

C++

#include <iostream>
using std::cout;
using std::cin;
using std::endl;

int array_size;
int sum = 0;

int main(){
    cout << "How many numbers ? " << endl;
    cin >> array_size;

    while (isdigit(array_size==false)){
        cout << "Input is not a number " << endl;
        break;
    } 
        int sum_array[array_size];
        for (int n=0; n < array_size; n++){
            cout << "Number " << n+1 << " is : ";
            cin >> sum_array[n];
            sum+=sum_array[n];
        }
        cout << "Sum of numbers are : " << sum << endl;
    return 0;
}

Here's the result when I tried to input integer and non-integer :

lock@lock ./a.out     
How many numbers ? 3
Number 1 is : 1
Number 2 is : 2
Number 3 is : 3
Sum of numbers are : 6

lock@lock ./a.out 
How many numbers ? a
Sum of numbers are : 0

See that in the second case when I put a non-integer, my line

cout << "Input is not a number " << endl;

is not being executed.

Now if I changed my condition with while (!isdigit(array_size)), here's the result :

lock@lock ./a.out 
How many numbers ? 3
Input is not a number 
Number 1 is : 1
Number 2 is : 2
Number 3 is : 3
Sum of numbers are : 6

lock@lock ./a.out 
How many numbers ? a
Input is not a number 
Sum of numbers are : 0

Apparently the line

cout << "Input is not a number " << endl;

will be executed no mater what the input (even if it's integer). What did I miss here ? How did "!" different from "false".

Posted 24-Sep-20 14:21pm

lock&_lock

Updated 24-Sep-20 20:39pm

CPallini

v2

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KarstenK 25-Sep-20 5:48am

the code is confusing. You need to improve the check and the code flow on an error.

4 solutions

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k5054 · Answer 1 · 2020-09-24T15:39:00

Solution 1

You have:

C++

isdigit(array_size==false)

This should be

C++

isdigit(array_size) == false

Note the placement of the brackets.

Posted 24-Sep-20 15:39pm

k5054

Comments

lock&_lock 25-Sep-20 14:39pm

Thanks, I tried your advice, but the line "cout << "Input is not a number " << endl;" still gets printed even if I put integer as an input

Patrice T · Answer 2 · 2020-09-24T15:40:00

Solution 2

I suspect you want to replace this

C++

while (isdigit(array_size==false)){

with this

C++

while (isdigit(array_size)==false){

But the logic of the loop look rather confuse. May be it is not a loop that you want.

Posted 24-Sep-20 15:40pm

Patrice T

Updated 24-Sep-20 15:50pm

v2

Comments

lock&_lock 25-Sep-20 14:40pm

This is similar to Soution 1. I tried this, but the line "cout << "Input is not a number " << endl;" still gets printed even if I put integer as an input

Patrice T 25-Sep-20 14:41pm

Yes Solution 1 was posted 1 minute before mine.

CPallini · Answer 3 · 2020-09-24T20:39:00

The argument of isdigit is supposed to be a character, but you are passing it a int variable.
Assuming your code should really deal with numbers (that is integers) and not digits, you might re-write your program this way (note you don't need an array):

C++

#include <iostream>
#include <limits>

using std::cout;
using std::cin;
using std::endl;


int main()
{
  int size;
  int sum = 0;

  cout << "How many numbers ? " << endl;

  for (;;)
  {
    cin >> size;
    if ( cin ) break;
    cin.clear();
    cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    cout << "Input is not a number " << endl;
  }

  for (int n = 0; n < size; ++n)
  {
    int i;
    cout << "Number " << (n+1) << " is : ";
    cin >> i;
    sum += i ;
  }
  cout << "Sum of numbers is : " << sum << endl;
}

Shao Voon Wong · Answer 4 · 2020-09-24T17:36:00

Solution 3

isdigit() takes in ASCII character though the parameter type is integer. Just change the type of array_size from int to char should work.

C++

char array_size;
...
cin >> array_size;
...
while (isdigit(array_size)==false){...

or

C++

char array_size;
...
cin >> array_size;
...
while (!isdigit(array_size)){...

should work.

Posted 24-Sep-20 17:36pm

Shao Voon Wong

v2

Comments

Stefan_Lang 25-Sep-20 6:24am

Actually, the argument type of isdigit is int. See https://www.cplusplus.com/reference/cctype/isdigit/

Your code will work because char gets auto-promoted to int. But declaring array_size as int is correct.

CPallini 25-Sep-20 19:41pm

Yes, the argument of isdigit is an int, however the function returns non-zero if the argument is in the range 0x30-0x39 (that is '0'..'9'). It is not useful to check if cin actually got an integer value.

Understanding of while loop and the difference of "! " and "false" in C++

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