linux下的make命令使用问题

Question

root@xianshoulai-A780GM-A:~/bag/8188eu_USB_linux/driver/rtl8188eu# make
make ARCH=i386 CROSS_COMPILE= -C /lib/modules/3.11.0-20-generic/build M=/home/xianshoulai/bag/8188eu_USB_linux/driver/rtl8188eu modules
make[1]: Entering directory `/usr/src/linux-headers-3.11.0-20-generic'
CC [M] /home/xianshoulai/bag/8188eu_USB_linux/driver/rtl8188eu/core/rtw_cmd.o
In file included from /home/xianshoulai/bag/8188eu_USB_linux/driver/rtl8188eu/core/rtw_cmd.c:23:0:
/home/xianshoulai/bag/8188eu_USB_linux/driver/rtl8188eu/include/osdep_service.h: In function ‘thread_enter’:
/home/xianshoulai/bag/8188eu_USB_linux/driver/rtl8188eu/include/osdep_service.h:1397:2: error: implicit declaration of function ‘daemonize’ [-Werror=implicit-function-declaration]
daemonize("%s", name);
^
cc1: some warnings being treated as errors
make[2]: *** [/home/xianshoulai/bag/8188eu_USB_linux/driver/rtl8188eu/core/rtw_cmd.o] Error 1
make[1]: *** [_module_/home/xianshoulai/bag/8188eu_USB_linux/driver/rtl8188eu] Error 2
make[1]: Leaving directory `/usr/src/linux-headers-3.11.0-20-generic'
make: *** [modules] Error 2

Answer 1

打错了罢? 是make && make intall
&& 是 shell 的逻辑 and (和), 如果前面的命令执行失败,(true or flase)
退出状态不是 0 , 就不会执行后面的命令 ,
最后命令的退出状态会在 shell 默认变量 $? 找到, 例

nc10@your-5554c55be4 ~
$ ls
Mail/ doc/ hh* matrix/ sign.dat trash
News/ ee* junk numsorted* socket.awk* ttt.awk*
bin/ ex30.sh* lib/ pp* software/
crack_allword.txt hex2dec.awk lynx.cfg sign tmp/

nc10@your-5554c55be4 ~
$ test -f "pp" && echo "yes, you have it"
yes, you have it

nc10@your-5554c55be4 ~
$ echo $?
0

nc10@your-5554c55be4 ~
$ test -f "lllll" && echo "yes, you have it"

nc10@your-5554c55be4 ~
$ echo $?
1

nc10@your-5554c55be4 ~
$
第二次执行测试文件 lllll,没有这文件， test名令执行失败,
( 条件为假(flase)), 随后的命令就不执行

nc10@your-5554c55be4 ~
$ test -f "lllll" ; echo "yes, you have it"
yes, you have it

nc10@your-5554c55be4 ~
$

分开两个命令，那不管前面执行如何，随后的还是会执行

如果分成两 make ; make install, 那make编译有错误的话，还是会照样
install , 但安装的程序不能用哦.

Is that clear? :)
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