Altera_Forum
Honored Contributor
15 years agowr3_addr and wr4_addr
could someone explain me the following verilog example.
I don't understand why the wr4_addr is 22'h280000 instead 22'h200000+320*256+1, i think that it is a choice of the programmer or has someone other reasons? thanks // FIFO Write Side 3 .WR3_DATA({1'b0, sCCD2_G[4:0], sCCD2_B[9:0]}), .WR3(sCCD2_DVAL), .wr3_addr(22'h200000), .WR3_MAX_ADDR(22'h200000+320*256), .WR3_LENGTH(9'h100), .WR3_LOAD(!DLY_RST_0), .WR3_CLK(CCD2_PIXCLK), .WR3_FULL(Full[2]), // FIFO Write Side 4 .WR4_DATA( {1'b0, sCCD2_G[9:5], sCCD2_R[9:0]}), .WR4(sCCD2_DVAL), .wr4_addr(22'h280000), .WR4_MAX_ADDR(22'h280000+320*256), .WR4_LENGTH(9'h100), .WR4_LOAD(!DLY_RST_0), .WR4_CLK(CCD2_PIXCLK), .WR4_FULL(Full[3]),