更多“one of the the four common features of spoken speaking is taking short cuts, e.g.incomplete sentences. () ”相关问题
  • 第1题:

    How many approaches are mentioned to define a narrative?

    A.One.

    B.Two.

    C.Three.

    D.Four.


    正确答案:C
    解析:细节理解题。由文章第二段中“Another”、“A third”可知定义故事的方法共有三种。

  • 第2题:

    下列程序的运行结果是( )。 SET EXACT ON s="ni"+SPACE(2) IF S=”ni” IF S=”ni” ?"one" ELSE ?"two" END IF ELSE IF S="ni" ?"three" ELSE ?"four" END IF END IF RETURN

    A.one

    B.two

    C.three

    D.four


    正确答案:C
    C。【解析】用==比较两个字符串时,当两个字符串完圣相同时,运算结果是逻辑真.T.。用=比较两个字符串时,运算结果与SETEXACTONIOFF的设置有关:0N先在较短的字符串的尾部加上若干个空格,使两个字符串的长度相等,然后进行精确比较;当处于0FF状态时,只要右边字符串与左边字符串的前面部分内容相匹配,即可得到逻辑真.T.。所以本题运行结果为three。

  • 第3题:

    How many morphemes does the word "impossible" consist of?

    A.One.
    B.Two.
    C.Three.
    D.Four.

    答案:C
    解析:
    考查词素知识。impossible这个单词包含三个语素,分别是前缀im-,词根possi和后缀-ble。故选C。

  • 第4题:

    下列语句能给数组赋值,而不使用for循环的是

    A.myArray{[1]="One";[2]="Two";[3]="Three";}

    B.String s[5]=new String[] {"Zero","One","Two","Three","Four"};

    C.String s[]=new String[] {"Zero","One","Two","Three","Four"};

    D.String s[]=new String[]= {"Zero","One","Two","Three","Four"};


    正确答案:C
    解析:字符串数组赋初值的方法有两种,一种是如选项C一样初始化。另外一种是先为每个数组元素分配引用空间,再为每个数组元素分配空间并赋初值。例如还可做如下赋值:
      string s[]=new String[5];
      s[0]="Zero";
      s[1]="One";
      s[2]="Two";
      s[3]="Three";
      s[4]="Four";

  • 第5题:

    ( 14 )下列程序的运行结果是

    SET EXACT ON

    s="ni"+SPACE(2)

    IF s=="ni"

    IF s="ni"

    ? "one"

    ELSE

    ? "two"

    ENDIF

    ELSE

    IF s="ni"

    ? "three"

    ELSE

    ? "four"

    ENDIF

    ENDIF

    RETURN

    A) one

    B) two

    C)three

    D)four


    正确答案:C

  • 第6题:

    下列给字符串二维数组进行赋值的语句中,错误的是()。

    A.String s[ ] [ ] = new String [ ] [ ] { { “One “ , “ Two “ }, { “ Three “ , “ Four “ } } ;

    B.String s[ ] [ ] = { { “ One “ , “Two “},{ “ Three “ , “ Four “ } } ;

    C.String s[ ] [ ] = new String [ ] [ ] { { “Zero”} , { “ One ” , “Two” , “ Three” , “ Four” } } ;

    D.String s[ 2] [2 ] = { { “ One ” , “Two”},{“ Three” , “ Four” } } ;


    String s[ 2] [2 ] = { { “ One ” , “Two”},{“ Three” , “ Four” } } ;