adding unsigned int to int
Solution 1
When an unsigned int
and an int
are added together, the int
is first converted to unsigned int
before the addition takes place (and the result is also an unsigned int
).
-1, while being the first negative number, is actually equivalent to the largest unsigned number - that is, (unsigned int) -1 === UINT_MAX
.
-2 in unsigned form is UINT_MAX - 1
, and so on, so -40 === UINT_MAX - 39 === 4294967256
(when using 32bit ints).
Of course, adding 4 then gives your answer:
4294967256 + 4 = 4294967260
.
This is a great quiz where you can learn some of the rules of integers in C (and similarly C++): http://blog.regehr.org/archives/721
Solution 2
Represent i and a in hexadecimal:
i = 4: 0x 0000 0004
a = -40: 0x FFFF FFD8
Following the implicit conversion rules of C++, a in a + i will be cast to unsigned int, that is, 4294967256. So a + i = 4294967260
Admin
Updated on June 14, 2022Comments
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Admin almost 2 years
#include <iostream> int main () { using namespace std; unsigned int i = 4; int a = -40; cout<<a+i<<endl; return 0; }
Executing this gives me 4294967260
I know there's a conversion taking place, from a signed int to unsigned int, but how and why this particular value? I noticed it's close to the sum of | 2147483647 | + 2147483647
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Sellorio almost 11 yearsCorrect. An alternative to this is to turn the 4 into an int and then add them, then if you want an unsigned int do an if block that if the result is >= 0 then it is converted to an unsigned int.
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almanegra over 8 yearsWhy does the
int
is converted tounsigned
and not the way around? -
wcochran over 5 yearsFor two's complement you end up with the same bit pattern -- the context of the result (which evidently is unsigned by default) determines whether the result is signed or unsigned.