C++: Is it safe to compare a 64bit integer with a 32bit integer?

Question

Assuming I have 2 variables:

uint64_t a = ...

uint32_t b = ...

Will comparing the integers yield the expected results, i.e. (a != b), or (b > a)?

http://stackoverflow.com/questions/6770258/how-do-promotion-rules-work-when-the-signedness-on-either-side-of-a-binary-opera — , Apr 11 '12 at 20:25

score 24 · Accepted Answer · answered Apr 11 '12 at 20:14

24

No problem. The compiler promotes the 32-bit to 64-bit before the comparison

answered Apr 11 '12 at 20:14

sizzzzlerz

Search for C++ integer promotion rules for more info. – bames53 Apr 11 '12 at 20:17
Yes, but if one is signed and the other is unsigned, then the comparison may not work. – Apr 11 '12 at 20:23
2

The question started "Assume I have .. uint64_t ... uint32_t". If we assume that, then neither is signed. – Robᵩ Apr 11 '12 at 20:30

score 8 · Answer 2 · answered Apr 11 '12 at 20:14

8

Short answer - yes. The 'smaller' is converted to bigger one before comparison.

answered Apr 11 '12 at 20:14

nothrow

2 Answers2