How to determinate the amount of digits in integer in C++? [closed]

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I'm trying to determinate the amount of digits in integer in C++ and so far I've tried the following methods:

1.

unsigned GetNumberOfDigits (unsigned i)
{
    return i > 0 ? (int) log10 ((double) i) + 1 : 1;
}

2.

int length_of_int(int input){
    int length = 0;
    while(input > 0){
        length++;
        input /= 10;
        if(input == 0)
            length++;
    }
    return length;
}

3.

int NumDigits(int x)  
{  
    x = abs(x);  
    return (x < 10 ? 1 :   
        (x < 100 ? 2 :   
        (x < 1000 ? 3 :   
        (x < 10000 ? 4 :   
        (x < 100000 ? 5 :   
        (x < 1000000 ? 6 :   
        (x < 10000000 ? 7 :  
        (x < 100000000 ? 8 :  
        (x < 1000000000 ? 9 :  
        10)))))))));  
}

And none works in my case, such as "000101" it has 6 digits, but it either says 4 or 3. Any help?

The purpose of this is checking a valid date, in format YYMMDD. I am aware that this type of format has Y2K-error but it's specified in the task it has to be that one.

Answer 1

You wrote And none works in my case, such as "000101" it has 6 digits, but it either says 4 or 3. Any help?

if the integer was 000101 then the first 3 zeros would get removed, it would become 101.
If it was a string you just count how much letters in the string.
Seems you want to represent binary I would use bit array for this
Edit: Okay it's not binary it's a date which should be stored into string to avoid this Y2K bug.

int count = strlen("000101");