如何计算浮点数的 div 和 mod？

发布于 2024-07-13 03:09:57 字数 864 浏览 7 评论 0原文

在 Perl 中，% 运算符似乎假定为整数。例如：

sub foo {
    my $n1 = shift;
    my $n2 = shift;
    print "perl's mod=" . $n1 % $n2, "\n";
    my $res = $n1 / $n2;
    my $t = int($res);
    print "my div=$t", "\n";
    $res = $res - $t;
    $res = $res * $n2;
    print "my mod=" . $res . "\n\n";
}   

foo( 3044.952963, 7.1 );
foo( 3044.952963, -7.1 );
foo( -3044.952963, 7.1 );
foo( -3044.952963, -7.1 );

现在如您所见

perl's mod=6
my div=428
my mod=6.15296300000033

perl's mod=-1
my div=-428
my mod=6.15296300000033

perl's mod=1
my div=-428
my mod=-6.15296300000033

perl's mod=-6
my div=428
my mod=-6.15296300000033

，我已经提出了一个用于计算 div 和 mod 的“解决方案”。然而，我不明白的是每个参数的符号应该对结果产生什么影响。 div 不是总是正数，即 n2 适合 n1 的次数吗？在这种情况下，算术应该如何工作？

原文

In Perl, the % operator seems to assume integers. For instance:

sub foo {
    my $n1 = shift;
    my $n2 = shift;
    print "perl's mod=" . $n1 % $n2, "\n";
    my $res = $n1 / $n2;
    my $t = int($res);
    print "my div=$t", "\n";
    $res = $res - $t;
    $res = $res * $n2;
    print "my mod=" . $res . "\n\n";
}   

foo( 3044.952963, 7.1 );
foo( 3044.952963, -7.1 );
foo( -3044.952963, 7.1 );
foo( -3044.952963, -7.1 );

gives

perl's mod=6
my div=428
my mod=6.15296300000033

perl's mod=-1
my div=-428
my mod=6.15296300000033

perl's mod=1
my div=-428
my mod=-6.15296300000033

perl's mod=-6
my div=428
my mod=-6.15296300000033

Now as you can see, I've come up with a "solution" already for calculating div and mod. However, what I don't understand is what effect the sign of each argument should have on the result. Wouldn't the div always be positive, being the number of times n2 fits into n1? How's the arithmetic supposed to work in this situation?

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无名指的心愿 2024-07-20 03:09:58

给定 a = qd + r，在计算 d 负值的余数时存在歧义。

例如：

表达式 −42 ÷ −5 可以表示为：
−42 = 9×(−5) + 3 或 −42 = 8×(−5) + (−2)。

所以余数要么是 3，要么是 −2。

有关更多信息： Wikipedia:Remainder "余数满足的不等式"

另外，输出mod / div 中负数的情况取决于软件语言的实现。请参阅维基百科：模运算（查看右侧的表格）

回复收藏 0 原文

彡翼 2024-07-20 03:09:57

标题提出一个问题，正文提出另一个问题。为了回答标题问题，就像在 C 中一样，% 运算符是整数模数，但有一个库例程“fmod”是浮点模数。

use POSIX "fmod";

sub foo {
    my $n1 = shift;
    my $n2 = shift;
    print "perl's fmod=" . fmod($n1,$n2), "\n";
    my $res = $n1 / $n2;
    my $t = int($res);
    print "my div=$t", "\n";
    $res = $res - $t;
    $res = $res * $n2;
    print "my mod=" . $res . "\n\n";
}

foo( 3044.952963, 7.1 );
foo( 3044.952963, -7.1 );
foo( -3044.952963, 7.1 );
foo( -3044.952963, -7.1 );

给出

perl's fmod=6.15296300000033
my div=428
my mod=6.15296300000033

perl's fmod=6.15296300000033
my div=-428
my mod=6.15296300000033

perl's fmod=-6.15296300000033
my div=-428
my mod=-6.15296300000033

perl's fmod=-6.15296300000033
my div=428
my mod=-6.15296300000033

The title asks one question, the body another. To answer the title question, just as in C, the % operator is an integer modulus, but there's a library routine "fmod" that's a floating point modulus.

use POSIX "fmod";

sub foo {
    my $n1 = shift;
    my $n2 = shift;
    print "perl's fmod=" . fmod($n1,$n2), "\n";
    my $res = $n1 / $n2;
    my $t = int($res);
    print "my div=$t", "\n";
    $res = $res - $t;
    $res = $res * $n2;
    print "my mod=" . $res . "\n\n";
}

foo( 3044.952963, 7.1 );
foo( 3044.952963, -7.1 );
foo( -3044.952963, 7.1 );
foo( -3044.952963, -7.1 );

gives

perl's fmod=6.15296300000033
my div=428
my mod=6.15296300000033

perl's fmod=6.15296300000033
my div=-428
my mod=6.15296300000033

perl's fmod=-6.15296300000033
my div=-428
my mod=-6.15296300000033

perl's fmod=-6.15296300000033
my div=428
my mod=-6.15296300000033

回复收藏 0 原文

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