Comparing int64_t and uint64_t

Question

Does anybody know why this code produces such an output? -1 >= 0!!!

[mahmood@tiger ~]$ cat t.cpp
#include <iostream>
#include <stdint.h>
int main()
{
  uint64_t i = 10;
  uint64_t j = 10;
  int64_t k = -1;
  std::cout << "k=" << k << " i-j=" << i-j;
  if (k >= i-j)
    std::cout << " --> yes k>=i-j\n";
  return 0;
}
[mahmood@tiger ~]$ g++ t.cpp
[mahmood@tiger ~]$ ./a.out
k=-1 i-j=0 --> yes k>=i-j
[mahmood@tiger ~]$

I know the types are different and a comparison need two similar types but at the end of the day, it is comparing -1 and 0. Isn't it?

Can I humbly offer a remedy? https://github.com/dbj-systems/dbj-laboratorium/tree/master/anything_but — Chef Gladiator, Mar 19 '19 at 06:37

score 3 · Accepted Answer · edited Jun 20 '20 at 09:12

if (k >= i-j)

both sides are converted to unsigned, so perhaps -1 is interpreted as 0xFFFFFFFFFFFFFFFF:

if (0xFFFFFFFFFFFFFFFF >= 0)

According to the standard (emphasis mine):

Expressions [expr]

Otherwise, if the operand that has unsigned integer type has rank greater than or equal to the rank of the type of the other operand, the operand with signed integer type shall be converted to the type of the operand with unsigned integer type.

score 2 · Answer 2 · answered Apr 19 '15 at 19:58

2

No. It's likely comparing 0xffffffffffffffff to 0. The signed variable gets promoted to the unsigned type before the comparison is made. Look up 'arithmetic conversions' in the standard for more details.

answered Apr 19 '15 at 19:58

Carl Norum

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I think it is called "converted", not "promoted". (Integral promotion is a slightly different thing.) – AlexD Apr 19 '15 at 20:12

Comparing int64_t and uint64_t

2 Answers2