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Home/ Questions/Q 7589837
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T20:15:36+00:00 2026-05-30T20:15:36+00:00

I am trying to read an unsigned int using cin as follows: #include <limits.h>

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I am trying to read an unsigned int using cin as follows:

#include <limits.h>
#include <iostream>

using namespace std;

int main(int argc, char* argv[])
{
    unsigned int number;

    // UINT_MAX = 4294967295
    cout << "Please enter a number between 0 and " << UINT_MAX << ":" << endl;

    cin >> number;

    // Check if the number is a valid unsigned integer
    if ((number < 0) || ((unsigned int)number > UINT_MAX))
    {
        cout << "Invalid number." << endl;
        return -1;
    }
    return 0;
}

However, whenever I enter a value greater than the upper limit of unsigned integer (UINT_MAX), the program displays 3435973836. How do I check if the input given by user falls between 0 to UINT_MAX?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T20:15:38+00:00Added an answer on May 30, 2026 at 8:15 pm

    Two things:

    • Checking if an unsigned integer is < 0 or > UINT_MAX is pointless, since it can never reach that value! Your compiler probably already complains with a warning like “comparison is always false due to limited range of type”.

    • The only solution I can think of is catching the input in a string, then use old-fashioned strtoul() which sets errno in case of overflow.

    I.e.:

    #include <stdlib.h>
    
    unsigned long number;
    std::string numbuf;
    cin >> numbuf;
    number = strtoul(numbuf.c_str(), 0, 10);
    if (ULONG_MAX == number && ERANGE == errno)
    {
        std::cerr << "Number too big!" << std::endl;
    }
    

    Note: strtoul returns an unsigned long; there’s no function strtou(), returning an unsigned int.

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