Is the behavior of % with negative operands defined in Perl 5?

Until recently (i.e. C99), the be开发者_高级运维havior of the modulo operator was implementation defined in C. Since Perl 5 is written in C, is it reliant on the behavior of the C compiler used to build it?

No, Perl 5 defines the modulo operator in perlop and even has tests to ensure it works as documented.

from perl/t/op/arith.t

tryeq $T++,  13 %  4, 1;
tryeq $T++, -13 %  4, 3;
tryeq $T++,  13 % -4, -3;
tryeq $T++, -13 % -4, -1;

However, if you use the integer pragma, you are at the mercies of the C compiler.

Perl implements its own modulo operator, but you can get to the one from your C compiler by using the integer pragma. perlop says

Note that when use integer is in scope, "%" gives you direct access to the modulo operator as implemented by your C compiler. This operator is not as well defined for negative operands, but it will execute faster.

That is, you have to be careful when you are using integer because the modulo might give you different answers.