代码搜索:dataIn

找到约 2,888 项符合「dataIn」的源代码

代码结果 2,888
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h dis.h

//****显示头文件****// #ifndef _dis_h #define _dis_h #ifndef unchar #define unchar unsigned char #define unint unsigned int #endif #include sbit readkey=P3^7; sbit datain=P1^3; sbit scl
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c cry12864.c

#include typedef unsigned char uchar; typedef unsigned int uint; extern const unsigned char shuzi_table[]; #define LCD_DataIn DDRB=0x00 //数据口方向设置为输入 #define LCD_DataOut DD
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vhd code.vhd

library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_arith.all; use ieee.std_logic_unsigned.all; entity code is port(datain:in std_logic_vector(7 downto 0); dataout1,dataout2,da
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vhd d_flipflop.vhd

library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_arith.all; use ieee.std_logic_unsigned.all; entity D_flipflop is port(ri,clk:in std_logic; datain :in std_logic_vector(7 downto 0);
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vhd d_flipflop.vhd

library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_arith.all; use ieee.std_logic_unsigned.all; entity D_flipflop is port(ri,clk:in std_logic; datain :in std_logic_vector(7 downto 0);
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java ch6_e6_30.java

public class ch6_e6_30 { public static void main(String args[]) { int[] dataIn = {49,45,80,11,18,106,55,251,91};//欲放入二叉树中的数据 BinaryTree searchTree = new BinaryTree(dataI
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txt 16×4bit的fifo设计.txt

16×4bit的FIFO设计。 LIBRARY IEEE; USE IEEE.STD_LOGIC_1164.ALL; USE IEEE.STD_LOGIC_UNSIGNED.ALL; ENTITY fifo2 IS PORT( datain:IN STD_LOGIC_VECTOR(3 DOWNTO 0); dataout:OUT STD_LOG
www.eeworm.com/read/402018/11543952

vhd d_flipflop.vhd

library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_arith.all; use ieee.std_logic_unsigned.all; entity D_flipflop is port(ri,clk:in std_logic; datain :in std_logic_vector(7 downto 0);
www.eeworm.com/read/258747/11845913

txt readme.txt

I/O inputs - Inputs : datain (8 bits) = serial input where message data is fed. If the message polynomial of form, d_nX^n + d_(n-1)X^(n-1) + ......etc., then d_n followed by d_(n-1) in the
www.eeworm.com/read/343298/11959556

asv bit_interleaver.asv

function [out]=Bit_interleaver(datain,length,rate); if isequal (rate,[0 0 0 0 0]) Ntds=2; Ncbps=100; Ntint=10; Ncyc=33; Ncbp6s=300; Nibp6s=100; elseif(rate==[0 0 0 0 1]