Question

这篇文章与我以前的帖子有关：

Weird data corruption in C++ code for unsigned long long

该帖子中提出的问题已得到解决，但现在我面临与截断无符号长long数据类型相关的其他问题。这发生在函数iPow（）中，用于将基数提升为指数的幂。 iPow（）的源代码在我之前的帖子中共享，但为了方便起见，下面转载：

 unsigned long long userinput::iPow(){

        while(this->revValue)
        {
                if (this->revValue & 1ULL)
                {
                        this->result *= this->value;
                }
                this->revValue >>= 1;
                this->value *= this->value;
        }
        return this->result;
}

在上面的函数中，value，revValue和result都是unsigned long long。类声明的源代码在上一篇文章中共享，但为方便起见，下面转载：

class userinput{

        public:
                userinput(unsigned long long);
                ~userinput();
                unsigned long long reverse ();
                unsigned long long raise_to_reverse ();

        private:
                userinput();
                userinput(const userinput&);
                void operator=(const userinput&);

        private:
                unsigned long long value;
                unsigned int numDigits;
                unsigned long long revValue;
                unsigned long long result;

        private:
                void calc_numDigits();
                unsigned long long iPow ();


};

我在gdb输出下面复制了this-＆gt;值的值，this-＆gt; revValue，以及在iPow（）循环的每次迭代中的this-＆gt;结果：

Enter a number between 0 and 99999 (exlusive):
1234

Breakpoint 1, userinput::iPow (this=0x7fffffffde10) at program.cpp:49
49      while(this->revValue)
Missing separate debuginfos, use: debuginfo-install glibc-2.15-56.fc17.x86_64 libgcc-4.7.0-5.fc17.x86_64 libstdc++-4.7.0-5.fc17.x86_64
(gdb) p this->revValue
$1 = 4321
(gdb) n
51          if (this->revValue & 1ULL)
(gdb) p this->revValue & 1ULL
$2 = 1
(gdb) n
53              this->result *= this->value;
(gdb) n
55          this->revValue >>= 1;
(gdb) p this->result
$3 = 1234
(gdb) n
56          this->value *= this->value;
(gdb) p this->revValue
$4 = 2160
(gdb) n
49      while(this->revValue)
(gdb) p this->value
$5 = 1522756
(gdb) n
51          if (this->revValue & 1ULL)
(gdb) p this->revValue & 1ULL
$6 = 0
(gdb) n
55          this->revValue >>= 1;
(gdb) n
56          this->value *= this->value;
(gdb) p this->revValue
$7 = 1080
(gdb) n
49      while(this->revValue)
(gdb) p this->value
$8 = 2318785835536
(gdb) n
51          if (this->revValue & 1ULL)
(gdb) p this->revValue & 1ULL
$9 = 0
(gdb) n
55          this->revValue >>= 1;
(gdb) n
56          this->value *= this->value;
(gdb) p this->revValue
$10 = 540
(gdb) n
49      while(this->revValue)
(gdb) p this->value
$11 = 3022197894083133696
(gdb) n
51          if (this->revValue & 1ULL)
(gdb) p this->revValue & 1ULL
$12 = 0
(gdb) n
55          this->revValue >>= 1;
(gdb) n
56          this->value *= this->value;
(gdb) p this->revValue
$13 = 270
(gdb) n
49      while(this->revValue)
(gdb) p this->value
$14 = 16658586050838462464
(gdb) n
51          if (this->revValue & 1ULL)
(gdb) p this->revValue & 1ULL
$15 = 0
(gdb) n
55          this->revValue >>= 1;
(gdb) n
56          this->value *= this->value;
(gdb) p this->revValue
$16 = 135
(gdb) n
49      while(this->revValue)
(gdb) p this->value
$17 = 8477297326610710528
(gdb) n
51          if (this->revValue & 1ULL)
(gdb) p this->revValue & 1ULL
$18 = 1
(gdb) n
53              this->result *= this->value;
(gdb) n
55          this->revValue >>= 1;
(gdb) p this->result
$19 = 1681011244301025280
(gdb) n
56          this->value *= this->value;
(gdb) p this->revValue
$20 = 67
(gdb) n
49      while(this->revValue)
(gdb) p this->value
$21 = 0
(gdb) n
51          if (this->revValue & 1ULL)
(gdb) p this->revValue & 1ULL
$22 = 1
(gdb) n
53              this->result *= this->value;
(gdb) n
55          this->revValue >>= 1;
(gdb) p this->result
$23 = 0
(gdb) n
56          this->value *= this->value;
(gdb) p this->revValue
$24 = 33
(gdb) n
49      while(this->revValue)
(gdb) p this->value
$25 = 0
(gdb)

如gdb输出中所示，此＆gt;值获得的最大值为 $ 17 = 8477297326610710528 ，之后该值在下一次迭代中截断为零。因此，由于循环逻辑，this-＆gt;结果的值也会截断为零。

是否存在可用于打印iPow（）结果的所有数字的数据类型，考虑范围0-99999（独占）？或者，如果没有可行的数据类型，我可以用什么方法将iPow（）的所有数字打印到stdout？

请分享您的想法。

TIA 维诺德

Answer 1

你的计算中有溢出。 unsigned long long在您的情况下有64位。 C ++规范提到无符号数的乘法使用模运算。也就是说，乘法的结果是正确的，直到2 ⁶⁴的倍数。

在你的情况下：

1681011244301025280是0x175426D200000000（为方便起见，我转换为十六进制）

乘以它自己：

0x175426D200000000 * 0x175426D200000000 =
(0x175426D2 * 2^32) * (0x175426D2 * 2^32) =
(0x175426D2 * 0x175426D2) * (2^32 * 2^32) =
(whatever) * 2^64) ≡ 0 (modulo 2^64)

因此，当64位不足以保持结果时，为什么你的大数乘以它给出0是完全可以理解的。

要消除溢出的可能性，请使用bigint类型（bigint是Internet搜索的关键字）。

bigint库的推荐问题是例如here

使用无符号long long时截断

1 个答案: