Why the unsigned integer produces a big output when the number stored through cin is negative?

Question

I want to know how the unsigned integer works.

#include <iostream>
using namespace std;

int main(){
    int a = 50;             //basic integer data type
    unsigned int b;         //unsigned means the variable only holds positive values
    cout << "How many hours do you work? ";
    cin >> b;
    cout << "Your wage is " << b << " pesos.";

}

But when I enter -5, the output is

Your wage is 4294967291 pesos.

And when I enter -1, the output is

Your wage is 4294967295 pesos.

Supposedly, unsigned integers hold positive values. How does this happen? And I only know basic C++. I don't understand bits.

Possible duplicate of [C/C++ unsigned integer overflow](https://stackoverflow.com/questions/16056758/c-c-unsigned-integer-overflow) — lubgr, Aug 14 '18 at 14:10
"_Supposedly, unsigned integers hold positive values_" Sure, they hold positive values. Your output is positive, so why do you think, that unsigned integers hold anything more, than positive values? — Algirdas Preidžius, Aug 14 '18 at 14:11
It's also entertaining, btw, when the `int` is negative (especially a near-zero negative number), against a properly read `unsigned int`. The promotions may surprise you. — WhozCraig, Aug 14 '18 at 14:23
@Eljay, good advice, but if you interpret the bits of 4294967291 as a 32-bit signed int, you get -5. Coincidence? or did the `>>` operator succeed (for some definition of "succeed.") — Solomon Slow, Aug 14 '18 at 14:23
@SomeWindowsUser That's pointless, and incurs a runtime cost. — George, Aug 14 '18 at 14:24
@jameslarge • implementation dependent partially succeeded, but actually failed. GIGO when ignoring failure checking, unless one has reason to believe the data is valid (but in this case, we know /a priori/ that we're throwing bad data at the input). — Eljay, Aug 14 '18 at 14:55
@lubgr, it is not a duplicate and it will never be since I did not understand integer security. I only understand basic C++ programming and never delved into bits of code since it wasn't taught. — Kenneth Ligutom, Aug 15 '18 at 12:42

score 3 · Accepted Answer · answered Aug 14 '18 at 14:23

Conversion from signed to unsigned integer is done by the following rule (§[conv.integral]):

A prvalue of an integer type can be converted to a prvalue of another integer type.
[...]

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2ⁿ where n is the number of bits used to represent the unsigned type).

In your case, unsigned is apparently a 32-bit type, so -5 is being reduced modulo 2³². So 2³² = 4,294,967,296 and 4,294,967,296 - 5 = 4,294,967,291.

Why the unsigned integer produces a big output when the number stored through cin is negative?

1 Answers1