Perl 5中是否定义了带有负操作数的％的行为? [英] Is the behavior of % with negative operands defined in Perl 5?

问题描述

直到最近(即C99)，模运算符的行为都是用C定义的实现.由于Perl 5是用C编写的，它是否依赖于用于构建它的C编译器的行为?

Until recently (i.e. C99), the behavior 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?

推荐答案

不，Perl 5在 perlop ，甚至进行了测试以确保其工作如文档所述.

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

来自perl/t/op/arith.t

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;

但是，如果您使用 integer 编译指示，您将受到C编译器.

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

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