• 2022-06-27
    下列代码描述中,不能产生时序逻辑的( )
    A: always (*)begainif (a&b) rega=c;elserega=0;end
    B: always (*)begainif (a&b) rega=c;y=rega;end
    C: always @(a)begainCase(a)2’b00: out=4’b0001;2’b01: out=4’b0010; 2’b10: out=4’b0100;endcaseend