The operator needs to execute a long-running shell script. How can the script be executed so the operator does not have to wait until it is complete to access the command prompt?()A. ./script.kshB. nohup ./script.kshC. run -b ./script.kshD. ./script.ksh >

题目
The operator needs to execute a long-running shell script. How can the script be executed so the operator does not have to wait until it is complete to access the command prompt?()

A. ./script.ksh

B. nohup ./script.ksh

C. run -b ./script.ksh

D. ./script.ksh > nohup.out


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  • 第1题:

    如果表达式x * y+z中,“*”是作为友元函数重载的,“+”是作为成员函数重载的,则该表达式还可为 ( )。

    A.operator+(operator$(X,Y),z)

    B.x.operator+(operator * (x,y),Z)

    C.Y.operator * (operator+(x,Y),z)

    D.x.operator+(operator{(X,Y))


    正确答案:A
    C++中用成员函数重载x*y为:x.oper-ator}(y),用友元函数重载x*Y为:(,perator*(x,Y),用成员函数重载x+Y为:x.operator+(Y),用友元函数重载x+Y为:operator+(x,Y)。

  • 第2题:

    如果表达式--X+y中,--是作为成员函数重载的,+是作为成员函数重载的,则该表达式还可为

    A.y.operator+(x.operator--(0))

    B.y,operator+(x.operator--( ))

    C.y.operator+(operator--t(x,0))

    D.operator+(x.operator- -t( ))


    正确答案:B
    解析:C++语言中用成员函数重载--x为:x.operator--(),用成员函数重载x+y为:x.operator+(y)。

  • 第3题:

    如果表达式++i*k时中的“++”和“*”都是重载的友元运算符,则采用运算符函数调用格式,该表达式还可表示为()。

    A.operator* (i. operator++(), k)

    B.operator* (operator++(i), k)

    C.i.operator++(). operator* (k)

    D.k. operator* (operator++(i))


    operator*(operator++(i),k)

  • 第4题:

    如果表达式--x/y中的“--”和“/”都是作为友元函数重载的运算符,采用运算符函数调用格式,该表达式还可表示( )。

    A.operator/(x. operator--(),y)

    B.operator/(operator--(x),y)

    C.x. operator--().operator/(y)

    D.y. operator/(operator--(x))


    正确答案:B

  • 第5题:

    如果表达式--x+y中,--是作为成员函数重载的,+是作为成员函数重载的,则该表达式还可为( )

    A.y.operator+(x.operator--(0))
    B.y.operator+(x.operator--())
    C.y.operator+(operator--(x,0))
    D.operator+(x.operator--())

    答案:B
    解析:

  • 第6题:

    如果表达式++i*k中的“++”和“*”都是重载的友元运算符,若采用运算符函数调用格式,则表达式还可以表示为()。

    A.operator*(i.operator++(),k)

    B.operator*(operator++(i),k)

    C.i.operator++().operator*(k)

    D.k.operator*(operator++(i))


    operator*(operator++(i),k)