int to unsigned int conversion

Question

I'm just amazed to know that I can't convert signed to unsigned int by casting!

int i = -62;
unsigned int j = (unsigned int)i;

I thought I already knew this since I started to use casts, but I can't do it!

Answer 1

You can convert an int to an unsigned int. The conversion is valid and well-defined.

Since the value is negative, UINT_MAX + 1 is added to it so that the value is a valid unsigned quantity. (Technically, 2^N is added to it, where N is the number of bits used to represent the unsigned type.)

In this case, since int on your platform has a width of 32 bits, 62 is subtracted from 2³², yielding 4,294,967,234.

Answer 2

Edit: As has been noted in the other answers, the standard actually guarantees that "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)". So even if your platform did not store signed ints as two's complement, the behavior would be the same.

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Apparently your signed integer -62 is stored in two's complement (Wikipedia) on your platform:

62 as a 32-bit integer written in binary is

0000 0000 0000 0000 0000 0000 0011 1110

To compute the two's complement (for storing -62), first invert all the bits

1111 1111 1111 1111 1111 1111 1100 0001

then add one

1111 1111 1111 1111 1111 1111 1100 0010

And if you interpret this as an unsigned 32-bit integer (as your computer will do if you cast it), you'll end up with 4294967234 :-)

Answer 3

This conversion is well defined and will yield the value UINT_MAX - 61. On a platform where unsigned int is a 32-bit type (most common platforms, these days), this is precisely the value that others are reporting. Other values are possible, however.

The actual language in the standard is

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