linux low memory请教 [问题点数:40分,结帖人guomsh]

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linux下overcommit_memory的问题
背景     公司的redis有时background save db不成功,通过log发现下面的告警,很可能由它引起的: [13223] 17 Mar 13:18:02.207 # WARNING overcommit_<em>memory</em> is set to 0! Background save may fail under <em>low</em> <em>memory</em> condition. To fix thi
Virtual Memory Low!
用Acess数据库存放2000多条图片记录.rnrn如果一次查询出的记录过多,程序就会很慢,而且系统会出现虚拟内存低的情况.rnrn有什么好的解决方法吗?rnrn有没有可能与数据库有关?rnrn
Available Memory Is Low
[img=http://hi.csdn.net/attachment/201107/30/52850_131203635609q9.jpg][/img]rn这是什么错误,怎么解决?
An Ultra Low Power Non-volatile Memory
Abstract-An ultra <em>low</em> power non-volatile <em>memory</em> is designed in a standard CMOS process for passive RFID tags. The <em>memory</em> can operate in a new <em>low</em> power operating scheme under a wide supply voltage and clock frequency range. In the charge pump circuit the threshold voltage effect of the switch transistor is almost eliminated and the pumping efficiency of the circuit is improved. An ultra <em>low</em> power 192-bit <em>memory</em> with a register array is implemented in a 0.18μM standard CMOS process. The measured results indicate that, for the supply voltage of 1.2 volts and the clock frequency of 780KHz, the current consumption of the <em>memory</em> is 1.8μA (3.6μA) at the read (write) rate of 1.3Mb/s (0.8Kb/s).
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请问如何解决???rn是不是硬件有问题
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我创建了数据库,然后试图写入数据时出现如上报错,部分放代码如下:rnrn rn case ctlSelectEvent:rnrn frmP=FrmGetActiveForm();rn SysManangeRogin(fld1,text1H);rn SysManageKey(fld2,fld3,text2H,text3H);rn rn EmployeeRecordP=DmNewRecord(dbP,&index,sizeof(Emp)); rn EmpP=MemHandleLock(EmployeeRecordP);rn DmWrite(EmpP,0,&Emp,sizeof(Emp));rn MemHandleUnlock(EmployeeRecordP);rn break;rn rnvoid SysManangeRogin(FieldPtr fld,MemHandle textH)rn rn if(FldGetTextLength(fld)>0)rn textH=FldGetTextHandle(fld);rn if(textH)rn Char *str;rn FldSetTextHandle(fld,NULL);rn str=MemHandleLock(textH);rn rn Emp.EmployeeName=str;rn rn //elsern
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android外行人,问个问题rnactivity不是在栈里管理嘛、那是不是<em>memory</em> <em>low</em>的时候,先从栈底开始杀?(有没有杀的优先级?)rnrn栈底下的被杀时候,当出栈到这个被杀的activity时,会发生什么 ?rnrn
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我在使用ralink的无线模组RT2571WF进行压力测试时(即持续上传数据)会导致<em>linux</em>内核out of <em>memory</em>,我使用的内核版本是2.6.14,arm9的cpu,以下是<em>linux</em>内核out of <em>memory</em>时串口打印的信息:rnrnoom-killer: gfp_mask=0xd0, order=0rnMem-info:rnDMA per-cpu:rncpu 0 hot: <em>low</em> 14, high 42, batch 7 used:22rncpu 0 cold: <em>low</em> 0, high 14, batch 7 used:0rnNormal per-cpu: emptyrnHighMem per-cpu: emptyrnFree pages: 1324kB (0kB HighMem)rnActive:419 inactive:601 dirty:0 writeback:0 unstable:0 free:331 slab:9115 mapped:451 pagetables:55rnDMA free:1324kB min:884kB <em>low</em>:1104kB high:1324kB active:1676kB inactive:2404kB present:49152kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnNormal free:0kB min:0kB <em>low</em>:0kB high:0kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnHighMem free:0kB min:128kB <em>low</em>:160kB high:192kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnDMA: 17*4kB 22*8kB 3*16kB 1*32kB 1*64kB 0*128kB 1*256kB 1*512kB 0*1024kB 0*2048kB 0*4096kB = 1156kBrnNormal: emptyrnHighMem: emptyrnSwap cache: add 0, delete 0, find 0/0, race 0+0rnFree swap = 0kBrnTotal swap = 0kBrnFree swap: 0kBrn12288 pages of RAMrn416 free pagesrn882 reserved pagesrn9114 slab pagesrn870 pages sharedrn0 pages swap cachedrnOut of Memory: Killed process 323 (subway).rnrnStop Sending, Switch Saving...rnoom-killer: gfp_mask=0xd2, order=0rnMem-info:rnDMA per-cpu:rncpu 0 hot: <em>low</em> 14, high 42, batch 7 used:36rncpu 0 cold: <em>low</em> 0, high 14, batch 7 used:6rnNormal per-cpu: emptyrnHighMem per-cpu: emptyrnFree pages: 1168kB (0kB HighMem)rnActive:381 inactive:557 dirty:0 writeback:0 unstable:0 free:292 slab:9219 mapped:379 pagetables:45rnDMA free:1168kB min:884kB <em>low</em>:1104kB high:1324kB active:1524kB inactive:2228kB present:49152kB pages_scanned:953 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnNormal free:0kB min:0kB <em>low</em>:0kB high:0kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnHighMem free:0kB min:128kB <em>low</em>:160kB high:192kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnDMA: 2*4kB 29*8kB 4*16kB 1*32kB 1*64kB 0*128kB 1*256kB 1*512kB 0*1024kB 0*2048kB 0*4096kB = 1168kBrnNormal: emptyrnHighMem: emptyrnSwap cache: add 0, delete 0, find 0/0, race 0+0rnFree swap = 0kBrnTotal swap = 0kBrnFree swap: 0kBrn12288 pages of RAMrn429 free pagesrn882 reserved pagesrn9219 slab pagesrn666 pages sharedrn0 pages swap cachedrnOut of Memory: Killed process 329 (subway).rnTRACE VE CH 0: Seq, 0rnrnStop Saving, Switch Sending...rnZSP_VENC_Stop...rnoom-killer: gfp_mask=0x601d2, order=0rnMem-info:rnDMA per-cpu:rncpu 0 hot: <em>low</em> 14, high 42, batch 7 used:39rncpu 0 cold: <em>low</em> 0, high 14, batch 7 used:8rnNormal per-cpu: emptyrnHighMem per-cpu: emptyrnFree pages: 1056kB (0kB HighMem)rnActive:387 inactive:559 dirty:0 writeback:0 unstable:0 free:257 slab:9278 mapped:385 pagetables:41rnDMA free:1028kB min:884kB <em>low</em>:1104kB high:1324kB active:1516kB inactive:2216kB present:49152kB pages_scanned:66 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnNormal free:0kB min:0kB <em>low</em>:0kB high:0kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnHighMem free:0kB min:128kB <em>low</em>:160kB high:192kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnDMA: 1*4kB 1*8kB 1*16kB 0*32kB 1*64kB 0*128kB 1*256kB 1*512kB 0*1024kB 0*2048kB 0*4096kB = 860kBrnNormal: emptyrnHighMem: emptyrnSwap cache: add 0, delete 0, find 0/0, race 0+0rnFree swap = 0kBrnTotal swap = 0kBrnFree swap: 0kBrn12288 pages of RAMrn350 free pagesrn882 reserved pagesrn9276 slab pagesrn647 pages sharedrn0 pages swap cachedrnOut of Memory: Killed process 347 (subway).rnoom-killer: gfp_mask=0x601d2, order=0rnMem-info:rnDMA per-cpu:rncpu 0 hot: <em>low</em> 14, high 42, batch 7 used:22rncpu 0 cold: <em>low</em> 0, high 14, batch 7 used:1rnNormal per-cpu: emptyrnHighMem per-cpu: emptyrnFree pages: 1180kB (0kB HighMem)rnActive:284 inactive:548 dirty:0 writeback:0 unstable:0 free:295 slab:9368 mapped:278 pagetables:31rnDMA free:1180kB min:884kB <em>low</em>:1104kB high:1324kB active:1132kB inactive:2196kB present:49152kB pages_scanned:1108 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnNormal free:0kB min:0kB <em>low</em>:0kB high:0kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnHighMem free:0kB min:128kB <em>low</em>:160kB high:192kB active:0kB inactive:0kB present:0kB pages_scanned:0 all_unreclaimable? norn<em>low</em>mem_reserve[]: 0 0 0rnDMA: 72*4kB 17*8kB 0*16kB 2*32kB 1*64kB 0*128kB 1*256kB 1*512kB 0*1024kB 0*2048kB 0*4096kB = 1320kBrnNormal: emptyrnHighMem: emptyrnSwap cache: add 0, delete 0, find 0/0, race 0+0rnFree swap = 0kBrnTotal swap = 0kBrnFree swap: 0kBrn12288 pages of RAMrn423 free pagesrn882 reserved pagesrn9368 slab pagesrn567 pages sharedrn0 pages swap cachedrnOut of Memory: Killed process 313 (subway).rnOut of Memory: Killed process 340 (subway).rnOut of Memory: Killed process 341 (subway).rnOut of Memory: Killed process 342 (subway).rnOut of Memory: Killed process 343 (subway).rnOut of Memory: Killed process 345 (subway).rnOut of Memory: Killed process 346 (subway).rnKilledrnAuto login as root ...rnrn请问哪位达人知道问题出在哪?请指点一下。rnrn非常感谢
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请问linux支持low pin count吗??
如果支持,在原代码的那个文件中,谢谢**
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跟大家讲讲cuda编程中遇到的很奇怪的问题:rn申请了一块constant内存存一个常数数组rn_device_ constant char fz[10]=rn'a','b','c','d','e',rn'f','g','h','i','j'rn这个数组在核函数中直接引用rn__global__ void myKernel(char* c)rnrn int count;rn //接下来是一些操作计算count rn c[1]=fz[count];rnrnrn然后在main函数中读c[1],在-emu模式下调试单步走是对的,但是在debug模式下却运行错误,读出来的值却不是fz[count],假如count明明是1,但fz[count]却不是'a',很不得其解????是不是constant <em>memory</em> 有什么我不知道的用法?<em>请教</em>各位高手了!rnrn
请教share memory的用法(很急)
我是个cuda新手,关于share <em>memory</em>的用法有几点不清楚:rnrn(1)假如我的kernel里要用到多个共享数组,我是否可以向下面这样声明:rnrn __shared__ int A[100][6];rnrn __shared__ int B[20];rnrn __shared__ int C[50];rnrn ,但是我好像在有些资料上看到说,同一个kernel里的每个share <em>memory</em>的起始地址都相同,需要计算偏移量,不能像上面那样做,我有些不确定,还请高手多多指教。rnrn(2)我写了一个cuda程序,编译通过,但运行时,出现错误,错误输出为:rn 1>------ 已启动生成: 项目: CUDA_3SAT, 配置: Release Win32 ------rn1>正在链接...rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 ___cudaUnregisterFatBinary@4rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 _cudaLaunch@4rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 _cudaSetupArgument@12rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 ___cudaRegisterFunction@40rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 ___cudaRegisterFatBinary@4rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 _cudaGetErrorString@4rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 _cudaMemcpy@16rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 _cudaConfigureCall@32rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 _cudaMalloc@8rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 __imp__cutCheckCmdLineFlag@12rn1>cuda_3sat.obj : error LNK2001: 无法解析的外部符号 _cudaFree@4rnrn,我曾尝试单步调试,不行,有人知道出现这种错误的原因吗?
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描述Linux kernel上NUMA的系统结构以及实现
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文章是Rincón-Mora写的十分经典的关于LDO是设计研究,学习LDO IC设计必看的宝典
AD621 Low Drift, Low Power Instrumentation Amplifier
AD621 Low Drift, Low Power Instrumentation AmplifierThe AD621 is an easy to use, <em>low</em> cost, <em>low</em> power, high accu- When operating from high source impedances, as in ECG and racy instrumentation amplifier which is ideally suited for a wide blood pressure monitors, the AD621 features the ideal combina- range of applications. Its unique combination of high perfor- tion of <em>low</em> noise and <em>low</em> input bias currents. Voltage noise is mance, small size and <em>low</em> power, outperforms discrete in amp specified as 9 nV/? Hz at 1 kHz and 0.28 ?V p-p from 0.1 Hz to implementations.
tomcat out of memory问题请教
昨天因为程序出错,导致内存溢出。我在setclasspath.bat中设置了JAVA_OPTS="-Xms512M -Xmx2000M",不知道是否有效,系统的物理内存是5G.不知道这样是否合理,另外刚才在自己机器上设置相同的参数,发现Xms不能超过一定的数量,比如我的机器是512M内存,其Xms不能超过64M,请问为什么?
low冒泡
import random def sort(t, p=1): if not isinstance(p, int): # 抛出异常 # return 强制中断,不往下执行 return TypeError(&quot;order的类型错误&quot;) # 参数3 st...
PoemArk - Lonely Low
PoemArk 20040420Lonely LowHave you ever been this <em>low</em>that you dont know where to go?Missing something you do long for,trying to get to heaven before they close the door?You walk s<em>low</em>ly along th
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请教masm的memory伪操作
我按照书上写rnS1 Segment rnS1 EndSrnS2 Segment MemoryrnS2 EndS rnrn编译器提示错误,why? 用Comman就正常rnrn
Low Power Design Low Power Design
Low Power Design Low Power Design
Low Latency
在集群中 Low Latency 应用,The operating systems community has ignored network latency for too long. In the past, speed-of-light delays in wide area networks and unoptimized network hardware have made sub-100μs round-trip times impossible. However, in the next few years datacenters will be deployed with <em>low</em>-latency Ethernet. Without the burden of propagation delays in the datacenter campus and network delays in the Ethernet devices, it will be up to us to finish the job and see this benefit through to applications. We argue that OS researchers must lead the charge in rearchitecting systems to push the boundaries of <em>low</em>latency datacenter communication. 5-10μs remote procedure calls are possible in the short term – two orders of magnitude better than today. In the long term, moving the network interface on to the CPU core will make 1μs times feasible.
the memory could not be memory
调用这个方法时出现了如下错误 请问是什么原因rnrnrn int status = 0;rnrn mysql_init(&mysql); rn rn tryrnrnrn if (mysql_real_connect(&mysql,"127.0.0.1","root","123456","sale",3306,0,0))rn rn rn status = 1;rnrn else rnrn status = 0; rnrn rn catch(CString ee)rnrnrnrn rnrnrn return status;rn
memory
Yi ge ji yi li xiao you xi
memory?
在看源代码的时候有这么一句:rnchar <em>memory</em> [MEMORY_SIZE];rn<em>memory</em>和[MEMORY_SIZE]之间有空格。这句话怎么理解?<em>memory</em>是关键字?作用?
video memory与AGP memory
请问高手,这两个中文应该怎么译,如果第一个叫显存第二个叫什么,并且有什么不同呀?
Memory
一个优秀的背单词软件 :-)
texture memory与global memory的区别
texture <em>memory</em>是个什么样的<em>memory</em>? 它与global <em>memory</em>有什么不同呢?比如定义和传输速度方面。它是不是就是global <em>memory</em>的一部分?望高手赐教!谢谢
Share Memory和Global Memory的对比。。。
我试验了以下两个kernel,一个用global一个share,但是为什么效果不明显,而且有的时候share要更慢呢?rnrn#include rn#include rnrn#define NUM 100000000rn#define BLOCKSIZE 100rnrn// __shared__ float sa[BLOCKSIZE];rnrn__global__ rnvoid mykernel(float* da , float* db , float* dc)rnrn __shared__ float sa[BLOCKSIZE];rn __shared__ float sb[BLOCKSIZE];rnrn int x = blockIdx.x * blockDim.x + threadIdx.x;rn dc[x] = 0;rn int idx = threadIdx.x;rn sa[idx] = da[x];rn sb[idx] = db[x];rn __syncthreads();rn for (int i = 0 ; i != blockDim.x ; ++ i)rn rn dc[x] += sa[i] * sb[i];rn rn for (int i = 0 ; i != blockDim.x ; ++ i)rn rn dc[x] += sa[blockIdx.x * blockDim.x + i] * sb[i];rn rnrnrn__global__ rnvoid mykernel1(float* da , float* db , float* dc)rnrn int x = blockIdx.x * blockDim.x + threadIdx.x;rn dc[x] = 0;rnrn for (int i = 0 ; i != blockDim.x ; ++ i)rn rn dc[x] += da[blockIdx.x * blockDim.x + i] * db[blockIdx.x * blockDim.x + i];rn rnrnrnrnint main()rnrn unsigned int timer = 0;rn cutilCheckError( cutCreateTimer( &timer));rn cutilCheckError( cutStartTimer( timer));rnrn float *ha , *hb , *hc;rn ha = (float*)malloc(sizeof(float) * NUM);rn hb = (float*)malloc(sizeof(float) * NUM);rn hc = (float*)malloc(sizeof(float) * NUM);rn for (int i = 0 ; i != NUM ; ++ i)rn rn ha[i] = 1;rn hb[i] = 2;rn rn float *da , *db , *dc;rn cudaMalloc(&da , sizeof(float) * NUM);rn cudaMalloc(&db , sizeof(float) * NUM);rn cudaMalloc(&dc , sizeof(float) * NUM);rnrn cudaMemcpy(da , ha , sizeof(float) * NUM , cudaMemcpyHostToDevice);rn cudaMemcpy(db , hb , sizeof(float) * NUM , cudaMemcpyHostToDevice);rnrn dim3 dimBlock(BLOCKSIZE , 1);rn dim3 dimGrid((NUM + dimBlock.x) / dimBlock.x , 1);rnrn mykernel1<<>>(da , db , dc);rnrn cudaMemcpy(hc , dc , sizeof(float) * NUM , cudaMemcpyDeviceToHost);rn cudaMemcpy(ha , da , sizeof(float) * NUM , cudaMemcpyDeviceToHost);rn cudaMemcpy(hb , db , sizeof(float) * NUM , cudaMemcpyDeviceToHost);rnrn cutilCheckError( cutStopTimer( timer));rn printf( "Processing time: %f (ms)\n", cutGetTimerValue( timer));rn cutilCheckError( cutDeleteTimer( timer));rnrn free(ha);rn free(hb);rn free(hc);rnrn cudaFree(da);rn cudaFree(db);rn cudaFree(dc);rn rn return 0;rn
关于Linux的缓存内存 Cache Memory详解
  转自:http://www.ha97.com/4337.html   Linux与Win的内存管理不同,会尽量缓存内存以提高读写性能,通常叫做Cache Memory。有时候你会发现没有什么程序在运行,但是使用top或free命令看到可用内存free项会很少,此时查看系统的 /proc/meminfo 文件,会发现有一项 Cached Memory: 输入cat /proc/m...
Linux的Memory used不断增加是为什么?
用top看只有一个占用1G内存的程序在运行,从top上看它的内存使用很稳定,物理内存不超过1G,虚拟内存不超过1.5G。但是这台一共有32G内存的机器的内存使用不断增加是为什么?或者说有没有什么工具可以完整检测出系统资源的使用?Linux的版本是kernel 2.6.9, Red Hat 4
‍Linux性能检查(二)Memory和IO
Linux性能检查(二)Memory和IO 内存包括物理内存和虚拟内存,虚拟内存(Virtual Memory)把计算机的内存空间扩展到硬盘,物理内存(RAM)和硬盘的一部分(SWAP)组合在一起作为虚拟内存为计算机提供连贯的虚拟内存空间。 好处是内存变多了,缺点是硬盘读写速度比内存慢很多,并且RAM和SWAP之间的交换增加了系统负担。 在操作系统里,虚拟内存被分成页,在X86系统上...
linux共享内存(share memory)与直接文件读取的区别
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linux环境下奇怪的malloc: memory corruption错误
小弟在<em>linux</em>下调试一个日志程序,发生以下错误:rnrn*** glibc detected *** ./log: malloc(): <em>memory</em> corruption: [color=#FF0000]0x08b23a58[/color] ***rn======= Backtrace: =========rn/lib/i686/nosegneg/libc.so.6[0xce4293]rn/lib/i686/nosegneg/libc.so.6(__libc_malloc+0x85)[0xce5c25]rn./log[0x804a767]rn./log[0x804acb6]rn./log[0x804ae9c]rn/lib/i686/nosegneg/libpthread.so.0[0xdda322]rn/lib/i686/nosegneg/libc.so.6(clone+0x5e)[0xd4d9ce]rn======= Memory map: ========rnrn该错误只在运行的内存以a58为结尾地址的时候才会出现,其他地址情况下运行正常。望各大虾赐教!rnrn我的环境为:Red Hat 4.1.2-12
回顾.Linux性能监控——CPU、Memory、IO、Network
Linux性能监控——CPU、Memory、IO、Network
the linux driver probe failed but it reside in memory still.
Hi all, I has a driver for i2c device, when I register it by insmod i2c_device.ko command, and it failed at probe function,then I return -1, but the i2c_device driver reside in <em>memory</em> still use the lsmod command to see it. now I want to the driver to exit <em>memory</em>, when it probe failed, so how I do it ? thanks!~~~~
Linux IPC——简单应用 共享内存(Share Memory)
目录   一:共享内存 1.1定义 1.2优缺点 1.2.1 优点 1.2.2缺点 1.3共享内存结构维护 1.4共享内存的通信原理 二:相关函数接口 2.1 Linux命令 2.1.1查看系统中的共享储存段  2.1.2删除系统中的共享储存段   2.2 函数 2.2.1shmget()创建共享内存函数 2.2.2shmctl()销毁等 函数 2.2.3shmat...
查看Linux中User对Memory的使用情况
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清华大学某 linux 驱动开发实验报告 memory
清华大学某 <em>linux</em> 驱动开发实验报告 <em>memory</em> 清华大学某 <em>linux</em> 驱动开发实验报告 <em>memory</em> 清华大学某 <em>linux</em> 驱动开发实验报告 <em>memory</em>
4G内存 安装linux cannot fit into memory
我机了内存是4G,硬盘安装Redhat Enterprise 5,在加载initrd.img 时出现如题目所示错误,详细过程如下rnrnroot (hd0,0)rnkernel (hd0,0)/vmlinuzrninitrd (hd0,0)/initrd.img rnrnError:Seleted item cannot fit into <em>memory</em>rnrn在网上找了下知道是因为内存大于2G了,initrd.img不能读进内存。但都没有找到解决的办法。rn大侠们帮帮忙呀。如休解决。
Code Commentary On The Linux Virtual Memory Manager
Code Commentary On The Linux Virtual Memory Manager Code Commentary On The Linux Virtual Memory Manager
Shared Memory Communications for Linux on IBM Z
Shared Memory Communications for Linux on IBM Z — Jing Zhang KVM on IBM Z Development 2018内核开发者大会
谁能详细谈谈system memory,video memory,AGP memory的区别?
system <em>memory</em>当然是系统内存了,那么vido <em>memory</em>和AGP <em>memory</em>分别指哪部分?rnrn为什么有的时候说显存快,有的时候又说系统内存快?rnrn请说说其各自特点,在什么情况下应该将数据放到哪里面,谢谢!
LOW EMI PCB DESIGN
INTEL资料,很实用,层叠、出线都有详尽的说明
low carbon zones
介绍低碳经济的最新中欧合作项目,让我们软件人员了解国家和国际上的合作前沿,从而拓展新的应用领域
DecisionCube Capacity is low
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七种qsort排序方法~~~下载
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相关热词 c#检测非法字符 c#双屏截图 c#中怎么关闭线程 c# 显示服务器上的图片 api嵌入窗口 c# c# 控制网页 c# encrypt c#微信网页版登录 c# login 居中 c# 考试软件
我们是很有底线的