如何以十进制格式打印64位整数值 [英] How to print 64-bit integer value in Decimal

问题描述

使用32位处理器以十进制格式打印64位整数值的方法应该是什么？



最近被问及面试。

What should be the approach to print a 64 bit Integer value in Decimals with a 32-bit processor ?

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推荐答案

为简单起见，假设你有一个 8位唯一的处理器和想要显示 16位无符号数。



假设 n 是输入数字，算法可以是（伪代码）：

In the sake of simplicity, suppose you have a 8-bit-only processor and want to show a 16 bit unsigned number.

Assuming n is the input number, the algorithm could be (pseudo code):

1. k = 0
2. 如果n = 0退出
3. r [k] =提醒n div 10
4. n = n div 10
5. k = k + 1
6. goto 2

（其中 div 是整数除法）



最后，数组 r 包含数字的十进制表示的数字（反向）。



当然在给定算法中存在问题：我们必须计算 16 位数的整数除法（和提醒）用仅8位处理器！



我们可以像手工一​​样执行此操作，但是时间使用4位换句话说，使用数字的十六进制表示。



让我们尝试在一个例子中做到这一点。

即我们试图将 n = 51728 除以 10 ，即 hexadecimalese ， 0xCA10 by 0xA 。

0xCA 是最有意义的字节， 0x10 是最不重要的字节。

第一步我们除 0xCA 的高位半字节，即 0xC by 0xA ，获取 0x1 作为提示， 0x2 作为提醒。

然后我们添加这样的提醒 0xCA 的最低半字节（即 0xA ），以获得 0x2A 并重复这个过程：

(where div is the integer division)

At the end, the array r contains the digits of the decimal representation of the number (in reverse order).

Of course there is a problem, in the given algorithm: we have to compute the integer division (and the reminder) of a 16 bit number with a 8-bit-only processor!

We could perform this operation like we do by hand, but using 4 bits at time, in other word using the hexadecimal representation of the numbers.

Let's try to do it in an example.
Namely we try to divide n = 51728 by 10, that is, in hexadecimalese, 0xCA10 by 0xA.
0xCA is the most significative byte and 0x10 is the least significative byte.
In the first step we divide the upper nibble of 0xCA, that is 0xC by 0xA, obtaining 0x1 as quotient and 0x2 as reminder.
Then we add such reminder to the lowest nibble of 0xCA (namely 0xA) in order obtain 0x2A and repeat the process:

// almost all hexadecimal numbers, here
 CA10            |A
                 ---
 2A               1           (C/A is 12/10, that is quotient 1, reminder 2)
  21               4          (2A/A is 42/10, that is quotient 4, reminder 2)
   30               3         (21/A is 33/10, that is quotient 3, reminder 3)
    6                4        (30/A is 48/10, that is quotient 4, reminder 8)

因此我们获得了商，即 0x1434 （5172十进制）和提醒 8 ，一次最多使用一个字节 。换句话说，我们的仅8位处理器可以执行它。

hence we have obtained the quotient, namely 0x1434 (5172 in decimal) and the reminder 8, always working with, at most, a byte at time. In other words, our 8-bit-only processor can perform it.

你可以尝试

You can try

printf("%I64d",c64);

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