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Home/ Questions/Q 7956915
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T04:08:19+00:00 2026-06-04T04:08:19+00:00

look at this code: void main () { int i = -1; unsigned u

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look at this code:

void main ()
{
int i = -1;
unsigned u = 1;

cout << u + i;
}

the addition of the u (unsigned) and i (signed), so i must be converted to the unsigned type, so it should be interpreted ( (2 ^ 32) – 1 ) and the expression should change from:
-1 + 1 to ( (2 ^ 32) – 1 ) + 1 but when i run the code it results to 0
why?

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  1. Editorial Team
    Editorial Team
    2026-06-04T04:08:21+00:00Added an answer on June 4, 2026 at 4:08 am

    -1 in an unsigned representation of the largest possible number unsigned can hold (UINT_MAX).

    Adding 1 to this wraps over due to the properties of unsigned, thus equaling 0.

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