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15 years agomapping a small table to a LUT instead of memory registers
I would like to map a table to a LUT simialr to the attached example in vhdl, but Quartus synthesis does not map the table in the attached verilog to a LUT, instead it uses memory registers.
Merely synthesizing the two equivalent modules in VHDL and verilog, shows the following synthesis reports. Apparently, there is lack of initialization option in verilog compared to vhdl that is causing this mapping to registers in verilog compared to LUTs in VHDL. What else can I do to map the table in verilog to LUT, other than the initial block, that is not working as desired. Thanks in advance, Fasahat --------------------------------------- VHDL Synthesis ------------------------------- Analysis & Synthesis Status Successful - Fri May 14 15:46:56 2010 Quartus II Version 9.0 Build 235 06/17/2009 SP 2 SJ Full Version Revision Name s_box Top-level Entity Name sbox Family Stratix II logic utilization n/acombinational aluts 40
dedicated logic registers 8 Total registers 8 Total pins 18 Total virtual pins 0 total block memory bits 0 DSP block 9-bit elements 0 Total PLLs 0 Total DLLs 0 ------------------------------ VERILOG Synthesis -------------------------------------------------- Analysis & Synthesis Status Successful - Fri May 14 15:48:33 2010 Quartus II Version 9.0 Build 235 06/17/2009 SP 2 SJ Full Version Revision Name s_box Top-level Entity Name s_box_rom2 Family Stratix II Logic utilization N/A combinational aluts 0
dedicated logic registers 0 Total registers 0 Total pins 17 Total virtual pins 0 total block memory bits 2,048 DSP block 9-bit elements 0 Total PLLs 0 Total DLLs 0