1

嗨，我得到了这个除法算法来显示整数和浮点值。如何获得 MAX_REM？它应该是我们的字符将被存储的缓冲区的大小，所以大小必须是数字的#，但我不知道如何得到它。谢谢！

void divisionAlg(unsigned int value)
{

    int MAX_BASE=10;
    const char *charTable = {"0123456789ABCDEF"};  // lookup table for converting remainders
    char rembuf[MAX_REM + 1];   // holds remainder(s) and provision null at the end
    int  index;                 // 
    int  i;                     // loop variable

    unsigned int rem;           // remainder
    unsigned int base;          // we'll be using base 10 
    ssize_t numWritten;         // holds number of bytes written from write() system call

    base = 10;

    // validate base
    if (base < 2 || base > MAX_BASE)
        err_sys("oops, the base is wrong");

    // For some reason, every time this method is called after the initial call, rembuf
    // is magically filled with a bunch of garbage; this just sets everything to null.
    // NOTE: memset() wasn't working either, so I have to use a stupid for-loop
    for (i=0; i<MAX_REM; i++)
        rembuf[i] = '\0';

    rembuf[MAX_REM] = 0;    // set last element to zero
    index = MAX_REM;        // start at the end of rembuf when adding in the remainders

    do
    {
        // calculate remainder and divide valueBuff by the base
        rem = value % base;
        value /= base;

        // convert remainder into ASCII value via lookup table and store in buffer
        index--;
        rembuf[index] = charTable[rem];

    } while (value != 0);

    // display value
    if ((numWritten = write(STDOUT_FILENO, rembuf, MAX_REM + 1)) == -1)
        err_sys("something went wrong with the write");

} // end of divisionAlg()
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1 回答 1

2

计算一个数字需要多少位数的计算方法是：

digits = floor(log(number)/log(base))+1;

但是，在这种情况下，我可能只是假设最坏的情况，因为它不超过 32，并且计算它会“昂贵”。所以只需#define MAX_REM 32, 然后跟踪你实际输入的数字rembuf（你已经有了index，所以它真的没有额外的成本）。您显然还需要计算要写出的字节数，但不需要任何特殊的数学运算。

于 2013-01-30T17:41:02.150 回答