?!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 把CE序的int main(void)Ҏstatic int main(void)会怎样呢?-南软g开发|׃软g开?南赢dU技软g开发公?/title> <meta name="keywords" content="把CE序的int main(void)Ҏstatic int main(void)会怎样呢?"/> <meta name="description" content="如题Q把CE序中的d数intmain(void)Ҏstaticintmain(void)会怎么样呢Q?nbsp;比如?nbsp;复制代码#include<stdio.h> intmain(void){  printf("Hi\n");  return0;}复制代码修改为: 复制代码#include<stdio.h> staticintmain"/> <link href="/css/style.css" rel="stylesheet" type="text/css" /> <meta http-equiv="Cache-Control" content="no-transform" /> <meta http-equiv="Cache-Control" content="no-siteapp" /> </head> <body> <div style="position:fixed;left:-9000px;top:-9000px;"><dfn id="orz4q"><noscript id="orz4q"><xmp id="orz4q"><output id="orz4q"></output></xmp></noscript></dfn><strong id="orz4q"><dl id="orz4q"></dl></strong><ol id="orz4q"><p id="orz4q"><nav id="orz4q"><pre id="orz4q"></pre></nav></p></ol><dfn 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href="/News/Technology/">软g开发技?/a> >> 览文章</div></div> <div class="youshi_f1" id="youshi_tdyx"> <div class="youshi01"> <h1 class="article_title">把CE序的int main(void)Ҏstatic int main(void)会怎样呢?</h1> <div class="article_author">d旉Q?016-12-20 17:26:18 文章作者:<a href="/">南软g开?/a> 览ơ数Q?Script Language="Javascript" Src="/item/GetHits.asp?Action=Count&GetFlag=0&m=1&ID=3077"></Script></div> <div class="article_main"><div id="MyContent"><p style="margin-top: 0px; margin-bottom: 0px; padding: 15px 0px 0px; line-height: 30px; font-family: 宋体; color: rgb(51, 51, 51); font-size: 13.3333px; white-space: normal; background-color: rgb(241, 241, 241);">如题Q把CE序中的d数int main(void)Ҏstatic int main(void)会怎么样呢Q?/p><p> </p><p>比如?/p><p> </p><p>复制代码</p><p>#include <stdio.h></p><p> </p><p>int main(void)</p><p>{</p><p>    printf("Hi\n");</p><p>    return 0;</p><p>}</p><p>复制代码</p><p>修改为:</p><p> </p><p>复制代码</p><p>#include <stdio.h></p><p> </p><p>static int main(void)</p><p>{</p><p>    printf("Hi\n");</p><p>    return 0;</p><p>}</p><p>复制代码</p><p> </p><p> </p><p>误者先自己想一惻I</p><p> </p><p> </p><p> </p><p> </p><p> </p><p>————————————————————分割线——————————————————?/p><p> </p><p>q个问题是我在看static关键字的时候提出来的?/p><p> </p><p>只要你了解static关键字会使标C符h内部链接QInternel LinkageQ属性,q且了解qCE序的编译链接流E,应该可以得出{案—?/p><p> </p><p>把CE序中的d数int main(void)Ҏstatic int main(void)会导致链接失败?/p><p> </p><p>可以验证一下:</p><p> </p><p>[zhanghaiba@Fedora code]$ gcc static_int_main.c</p><p>/usr/lib/gcc/i686-redhat-linux/4.4.5/../../../crt1.o: In function `_start':</p><p>(.text+0x18): undefined reference to `main'</p><p>collect2: ld returned 1 exit status</p><p> </p><p>如果换成gcc -c呢?</p><p> </p><p>[zhanghaiba@Fedora code]$ gcc -c static_int_main.c</p><p>[zhanghaiba@Fedora code]$</p><p> </p><p>可见换成gcc -c可以~译成功Q因为gcc -c只有预处理、编译和汇编阶段Q没有链接阶Dc?/p><p> </p><p> </p><p> </p><p>首先Q我们要了解一下Linux下GCC环境中CE序的编译链接流E—?/p><p> </p><p>~译CE序Q一般包括了C预处理阶DcC到汇~的~译阶段、汇~到目标文g的编译阶Dc目标文件的链接阶段?/p><p> </p><p>GCC支持下面几个命oQ我们可以观察到这些阶D:</p><p> </p><p>1Qgcc -v GCC.c</p><p> </p><p>~译时打印出ȝ~译程Q可以看C用了哪些~译工具。v是verboseQ冗长)的意思,卛_可能多的打印信息?/p><p> </p><p>2) gcc -E GCC.c</p><p> </p><p>把源文g用预处理器处理,可重定向输出到GCC.i文g再查?/p><p> </p><p>3Qgcc -S GCC.c</p><p> </p><p>把源文g用预处理器和~译器处理,自动输出同名的GCC.s文g</p><p> </p><p>4Qgcc -c GCC.c</p><p> </p><p>把源文g用预处理器、编译器和汇~器处理Q自动输出同?o文g</p><p> </p><p>5Qgcc GCC.c</p><p> </p><p>把源文g用预处理器、编译器、汇~器处理后,最后用链接器生成~省名ؓa.out的可执行文g</p><p> </p><p>Z么默认叫a.outQ因为早期编译ƈ没有链接器的概念Qa.out是汇~器直接生成的,a意ؓassembly。但需要澄清的是在C~译器中a.out都是由链接器生成?/p><p> </p><p>另外Q用选项-save-temps可以保留中间生成的文ӞC如下Q?nbsp;</p><p> </p><p>[zhanghaiba@Fedora code]$ ls | grep hi</p><p>hi.c</p><p>hi.i</p><p>hi.o</p><p>hi.s</p><p> </p><p> </p><p>我们再用gcc -v来观察ȝ~译程</p><p> </p><p>复制代码</p><p>[zhanghaiba@Fedora code]$ gcc -v hi.c</p><p>Using built-in specs.</p><p>Target: i686-redhat-linux</p><p>Configured with: ../configure --prefix=/usr --mandir=/usr/share/man --infodir=/usr/share/info --with-bugurl=http://bugzilla.redhat.com/bugzilla --enable-bootstrap --enable-shared --enable-threads=posix --enable-checking=release --with-system-zlib --enable-__cxa_atexit --disable-libunwind-exceptions --enable-gnu-unique-object --enable-languages=c,c++,objc,obj-c++,java,fortran,ada --enable-java-awt=gtk --disable-dssi --with-java-home=/usr/lib/jvm/java-1.5.0-gcj-1.5.0.0/jre --enable-libgcj-multifile --enable-java-maintainer-mode --with-ecj-jar=/usr/share/java/eclipse-ecj.jar --disable-libjava-multilib --with-ppl --with-cloog --with-tune=generic --with-arch=i686 --build=i686-redhat-linux</p><p>Thread model: posix</p><p>gcc version 4.4.5 20101112 (Red Hat 4.4.5-2) (GCC) </p><p>COLLECT_GCC_OPTIONS='-v' '-mtune=generic' '-march=i686'</p><p> /usr/libexec/gcc/i686-redhat-linux/4.4.5/cc1 -quiet -v hi.c -quiet -dumpbase hi.c -mtune=generic -march=i686 -auxbase hi -version -o /tmp/ccrwAICf.s</p><p>ignoring nonexistent directory "/usr/lib/gcc/i686-redhat-linux/4.4.5/include-fixed"</p><p>ignoring nonexistent directory "/usr/lib/gcc/i686-redhat-linux/4.4.5/../../../../i686-redhat-linux/include"</p><p>#include "..." search starts here:</p><p>#include <...> search starts here:</p><p> /usr/local/include</p><p> /usr/lib/gcc/i686-redhat-linux/4.4.5/include</p><p> /usr/include</p><p>End of search list.</p><p>GNU C (GCC) version 4.4.5 20101112 (Red Hat 4.4.5-2) (i686-redhat-linux)</p><p>    compiled by GNU C version 4.4.5 20101112 (Red Hat 4.4.5-2), GMP version 4.3.1, MPFR version 2.4.2.</p><p>GGC heuristics: --param ggc-min-expand=81 --param ggc-min-heapsize=95788</p><p>Compiler executable checksum: e892644090a9a7e8c330a388c51818dd</p><p>COLLECT_GCC_OPTIONS='-v' '-mtune=generic' '-march=i686'</p><p> as -V -Qy -o /tmp/cc1w7Hxi.o /tmp/ccrwAICf.s</p><p>GNU assembler version 2.20.51.0.2 (i686-redhat-linux) using BFD version version 2.20.51.0.2-15.fc13 20091009</p><p>COMPILER_PATH=/usr/libexec/gcc/i686-redhat-linux/4.4.5/:/usr/libexec/gcc/i686-redhat-linux/4.4.5/:/usr/libexec/gcc/i686-redhat-linux/:/usr/lib/gcc/i686-redhat-linux/4.4.5/:/usr/lib/gcc/i686-redhat-linux/:/usr/libexec/gcc/i686-redhat-linux/4.4.5/:/usr/libexec/gcc/i686-redhat-linux/:/usr/lib/gcc/i686-redhat-linux/4.4.5/:/usr/lib/gcc/i686-redhat-linux/</p><p>LIBRARY_PATH=/usr/lib/gcc/i686-redhat-linux/4.4.5/:/usr/lib/gcc/i686-redhat-linux/4.4.5/:/usr/lib/gcc/i686-redhat-linux/4.4.5/../../../:/lib/:/usr/lib/</p><p>COLLECT_GCC_OPTIONS='-v' '-mtune=generic' '-march=i686'</p><p> /usr/libexec/gcc/i686-redhat-linux/4.4.5/collect2 --no-add-needed --eh-frame-hdr --build-id -m elf_i386 --hash-style=gnu -dynamic-linker /lib/ld-linux.so.2 /usr/lib/gcc/i686-redhat-linux/4.4.5/../../../crt1.o /usr/lib/gcc/i686-redhat-linux/4.4.5/../../../crti.o /usr/lib/gcc/i686-redhat-linux/4.4.5/crtbegin.o -L/usr/lib/gcc/i686-redhat-linux/4.4.5 -L/usr/lib/gcc/i686-redhat-linux/4.4.5 -L/usr/lib/gcc/i686-redhat-linux/4.4.5/../../.. /tmp/cc1w7Hxi.o -lgcc --as-needed -lgcc_s --no-as-needed -lc -lgcc --as-needed -lgcc_s --no-as-needed /usr/lib/gcc/i686-redhat-linux/4.4.5/crtend.o /usr/lib/gcc/i686-redhat-linux/4.4.5/../../../crtn.o</p><p> </p><p>复制代码</p><p> </p><p> </p><p>注意U色加粗部分Q由l色文g生成U色文gQ—?/p><p> </p><p>Q?Qcc1是GCC~译环境中的C~译器,把C代码~译为汇~代码,输出?s文g</p><p> </p><p>Q?Qas是汇~器Q把汇编代码~译为目标文Ӟ输出?o文g</p><p> </p><p>Q?Qcollect2是GCC后期版本使用的链接器Q环境)Q其实是先调用GNU的链接器ld对目标文件进行链接,最后收集与E序初始化相关的信息Q构造程序的初始化结构?/p><p> </p><p>  ld是真正的链接器,对上一步的.o目标文g和其它需?o文g或静态链接库.a文g、动态链接库.so文gQ如解压C标准库libc.a中取出需要的printf.o文gQ,一起链接输Zؓa.out文g?/p><p> </p><p>  GCC后期版本使用了collect2来作为链接器Q其实是间接调用ld链接器?/p><p> </p><p>上面用到的工具中Qas是GNU自带的汇~器Qld是GNU自带的链接器Q它俩是GNU Binutils中最主要的二q制工具?/p><p> </p><p>其中Qld-linux.so.2是动态链接器。最后注?lc参数Ql表示链接Qc表示标准C库,?libc.a或libc.so?/p><p> </p><p> </p><p> </p><p>让我们回到问题本w—?/p><p> </p><p>main不是C语言的关键字Q但却是U定俗成的主函数名字Q不q它q不是程序执行的入口Q?/p><p> </p><p>CE序真正入口是_start全局W号Q由汇编实现的函敎ͼQ_start函数会调用库函数__libc_start_mainQ然后__libc_start_main再调用main函数</p><p> </p><p>我们知道main函数的声明无非两UŞ式,main函数的声明(mainW号Q其实是在crt1.o目标文g?/p><p> </p><p>通过nm工具可以查看crt1.o包括了哪些符?/p><p> </p><p>复制代码</p><p>[zhanghaiba@Fedora code]$ nm /usr/lib/crt1.o</p><p>00000000 R _IO_stdin_used</p><p>00000004 D __data_start</p><p>         U __libc_csu_fini</p><p>         U __libc_csu_init</p><p>         U __libc_start_main</p><p>00000044 R _fp_hw</p><p>00000020 T _start</p><p>00000004 W data_start</p><p>         U main</p><p>复制代码</p><p>crt1.o中已l有了mainW号Q但却是未定?U)的,所以需要我们来实现main函数Q即定义mainW号Q,最后通过链接器来链接Q这里称作符可析)</p><p> </p><p>如果把main函数定义为staticQ也是h内部链接(Internel Linkage)属性,则编译后的目标文件是局部符P当前文g可见Q?/p><p> </p><p>然而链接是不会对局部符号做W号解析的,只会Ҏ目标文g?rel.textD|指示链接全局的且未定义的W号Q即修改可重定位目标文gREL的符号地址Q?/p><p> </p><p>因此Q链接时mainW号找不到定义,q导致mainW号找不到具体实玎ͼ定义Q,造成链接p|</p><p> </p><p> </p><p> </p><p>我们再看~译p|的反馈信?/p><p> </p><p>/usr/lib/gcc/i686-redhat-linux/4.4.5/../../../crt1.o: In function `_start':</p><p>(.text+0x18): undefined reference to `main'</p><p>collect2: ld returned 1 exit status</p><p> </p><p>׃隄解了—?/p><p> </p><p>在函数_start中,引用了未定义的符号main</p><p> </p><p>collect2外壳Q链接器ldq回1标记退出状态(出错状态)</p><p><br/></p></div> </div> </div> </div> <div class="clear"></div> <div class="foot"> <div class="foot_menu"> <ul> <li><a href="/About/">关于我们</a></li> <li><a href="/Advantage/">开发优?/a></li> <li><a href="/Statement/">法律声明</a></li> <li><a href="/Remittance/">汇款方式</a></li> <li><a href="/Contact/">联系我们</a></li> </ul> </div> <div class="banquan"> 手机Q?8678812288 EQMail:1069706080@qq.com<br /> 地址Q山东省南市舜耕\泉城公园东门园内向北50c? 鲁ICP?7011972? 版权所?008Q?013 ׃赢d信息U技有限公司<script type="text/javascript"> 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