代码搜索:Process
找到约 10,000 项符合「Process」的源代码
代码结果 10,000
www.eeworm.com/read/494695/6360566
vhd 一个简单的uart.vhd
----------------------------------------------------------------
--
-- Copyright (c) 1992,1993,1994, Exemplar Logic Inc. All rights reserved.
--
-------------------------------------------------------
www.eeworm.com/read/494536/6377357
plg gsm.plg
礦ision2 Build Log
Project:
F:\GSM0610\gsm\gsm.uv2
Project File Date: 08/22/2008
Output:
Build target 'Target 1'
assembling STARTUP.A51...
co
www.eeworm.com/read/493331/6400919
rpt exp9.map.rpt
Analysis & Synthesis report for exp9
Tue May 19 08:57:11 2009
Version 5.1 Build 176 10/26/2005 SJ Full Version
---------------------
; Table of Contents ;
---------------------
1. Legal No
www.eeworm.com/read/493331/6400946
qmsg exp9.map.qmsg
{ "Info" "IQEXE_SEPARATOR" "" "Info: *******************************************************************" { } { } 3 0 "*******************************************************************" 0 0}
{ "I
www.eeworm.com/read/493099/6406395
cpp 车厢调度2.3ywc.cpp
#include
#include
#define MaxLen 100
struct snode{
int data[MaxLen];
int top;
}s;//定义一个栈指针
int n;//定义输入序列总个数
void
www.eeworm.com/read/492806/6409278
plg name.plg
礦ision3 Build Log
Project:
D:\实验\实验参考程序\c代码\LCD12864J\name.uv2
Project File Date: 05/09/2008
Output:
Build target 'Target 1'
compiling 12864J
www.eeworm.com/read/492901/6412083
cpp d10r1.cpp
#include
#include
#include
#include
double func(double x)
{
double t;
t=bessj0( x);
return t;
}
void main()
{
//program d10r1
www.eeworm.com/read/492926/6414147
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/492297/6420673
plg 12.plg
uVision3 Build Log
Project:
F:\wxq\饮水机\12.uv2
Project File Date: 04/29/2009
Output:
Build target 'Target 1'
compiling main.c...
compiling dr
www.eeworm.com/read/492075/6423950
txt pipe.txt
#include
#include
#include
#include
int main()
{
int fd[2], status;
int cld1_pid, cld2_pid, cld3_pid;
static int x1 = 0, x2 = 0, x