Today's response to the data access dilemma eschews the traditional archival approach in favor of emerging standards for what are described as the documents of the future: compound documents. Product users and developers hope the latest standards effort--

题目

Today's response to the data access dilemma eschews the traditional archival approach in favor of emerging standards for what are described as the documents of the future: compound documents. Product users and developers hope the latest standards effort--such as OpenDoc and Microsoft Corp.'s Object Linking and Embedding(OLE) will(26)data accessibility.OpenDoc is a standards group(27)a common look and feel throughout documents, even though different applications are used.OLE technologies allow Microsoft to(28)its various software products. Microsoft creates, governs and distributes the OLE capability, which is what most(29)it from the open OpenDoc approach. Thus, users can assemble(30)of varying application to create compound documents. With OLE users can, for example, embed an Excel chart in a Microsoft word report.

A.get

B.make

C.ensure

D.take


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  • 第1题:

    LTERRC连接建立原因有哪些

    A.MO (Mobile Originating) – signaling

    B.MO – data;

    C.MT (Mobile Terminating) – access;

    D.Emergency;

    E.highPriorityAccess;


    参考答案:A,B,C,D,E

  • 第2题:

    下列程序段的输出结果是()。 int data = 0; char k = 'a', p = 'f'; data = p - k; System.out.printlln(data);

    A.0

    B.a

    C.5

    D.f


    D 解析:a和f的ASCII值相差5。

  • 第3题:

    7、一个顺序栈S,其栈顶指针为top,则将元素e入栈的操作是()。

    A.S.data[top]=e;S.top++;

    B.S.top++;S.data[top]=e;

    C.S.data[top]=e;

    D.S.top=e;


    *S->top=e;S->top++;

  • 第4题:

    To the scientists' delight, their efforts have increased people's _______of the importance of saving water.

    A.awareness
    B.commitment
    C.agreement
    D.response

    答案:A
    解析:
    考查名词辨析。句意为“让科学家们高兴的是,他们的努力增加了人们的节约用水的”。awareness"意识”,commitment“承诺”,agreement“合同,同意”,response“回应”。根据题意,只有A项符合逻辑。故选A。

  • 第5题:

    17、下列程序段的输出结果是()。 int data = 0; char k = 'a', p = 'f'; data = p - k; System.out.printlln(data);

    A.0

    B.a

    C.5

    D.f


    20, y=30; boolean b;

  • 第6题:

    下列程序段的输出结果是()。 int data = 0; char k = 'a', p = 'f'; data = p - k; System.out.printlln(data);

    A.0

    B.5

    C.a

    D.f


    D D。【解析】a和f的ASCIl值相差5。