計算一個數(shù)的二進(jìn)制表示中有多少個1

這個階段的我好像很喜歡把玩code。拿到Code會想它是否能夠工作。先使用，再Trace，然后總結(jié)歸納。這些都是從Points C中摘錄的。不知道文章該該怎么分類。所以暫且歸為翻譯這一欄目。
    現(xiàn)在看來還是覺得這些簡短的Code很經(jīng)典。
    *********************************************方法一**********************************************************
    /**//*
    ** This function returns the number of 1-bits that appeared in
    ** the argument value.
    */
    int
    count_one_bits( unsigned value )
    ...{
     int ones;
     /**//*
     ** While the value still has some 1-bits in it ...
     */
     for( ones = 0; value != 0; value = value >> 1 )
     /**//*
     ** If the low-order bit is a 1, count it.
     */
     if( value % 2 != 0 )
     ones = ones + 1;
     return ones;
    }
    ******************************************方法二*******************************************
    /**//*
    ** This function returns the number of 1-bits that appeared in
    ** the argument value.
    */
    int
    count_one_bits( unsigned value )
    ...{
     int ones;
     /**//*
     ** While the value still has some 1-bits in it ...
     */
     for( ones = 0; value != 0; value >>= 1 )
     /**//*
     ** If the low-order bit is a 1, count it.
     */
     if( ( value & 1 ) != 0 )
     ones += 1;
     return ones;
    }