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Virtualenv不适用于RedHat(Virtualenv doesn't work on RedHat)

 我试图找出为什么我不能在RedHat上创建virtualenv。 这是我的命令行：  
python /opt/rh/python27/root/usr/bin/virtualenv-2.7 -p /opt/rh/python27/ /home/myuser/tpa/
 
 。  
/opt/rh/python27/root/usr/bin/virtualenv-2.7 : Is here to setup the version of virtualenv that I want to use

-p /opt/rh/python27/  : To setup the version of python I want to use

/home/myuser/tpa/ : My directory where the virtualenv will be stored
 
 但我得到以下内容：  
Traceback (most recent call last):
  File "/opt/rh/python27/root/usr/bin/virtualenv-2.7", line 5, in <module>
    from pkg_resources import load_entry_point
  File "/usr/lib/python2.6/site-packages/setuptools-0.6c11-py2.6.egg/pkg_resources.py", line 2603, in <module>
  File "/usr/lib/python2.6/site-packages/setuptools-0.6c11-py2.6.egg/pkg_resources.py", line 666, in require
  File "/usr/lib/python2.6/site-packages/setuptools-0.6c11-py2.6.egg/pkg_resources.py", line 565, in resolve
pkg_resources.DistributionNotFound: virtualenv==1.10.1
 
 你有什么主意吗？ 谢谢 

I'm trying to figure out why I cannot create a virtualenv on RedHat. Here's my command line : 
python /opt/rh/python27/root/usr/bin/virtualenv-2.7 -p /opt/rh/python27/ /home/myuser/tpa/
 
. 
/opt/rh/python27/root/usr/bin/virtualenv-2.7 : Is here to setup the version of virtualenv that I want to use

-p /opt/rh/python27/  : To setup the version of python I want to use

/home/myuser/tpa/ : My directory where the virtualenv will be stored
 
But I get the following :  
Traceback (most recent call last):
  File "/opt/rh/python27/root/usr/bin/virtualenv-2.7", line 5, in <module>
    from pkg_resources import load_entry_point
  File "/usr/lib/python2.6/site-packages/setuptools-0.6c11-py2.6.egg/pkg_resources.py", line 2603, in <module>
  File "/usr/lib/python2.6/site-packages/setuptools-0.6c11-py2.6.egg/pkg_resources.py", line 666, in require
  File "/usr/lib/python2.6/site-packages/setuptools-0.6c11-py2.6.egg/pkg_resources.py", line 565, in resolve
pkg_resources.DistributionNotFound: virtualenv==1.10.1
 
Do you have any idea? Thanks

原文：https://stackoverflow.com/questions/37530575

更新时间：2023-05-07 17:05

最满意答案

 我认为你的输出应该是“一些随机字符串”，因为你的两行代码什么都不做，请参阅注释。  
int main(){
   char token[] = "some random string";
   char c[23];  
   strcpy( c, token);
   strncpy(c, token, 5);  // Does nothing
   c[strlen(c)] = '\0';   // Does nothing
   cout<<c;
   return 0 ;
}
 
 如果要输出“some”，可以这样做：  
strncpy(c, token, 5);
c[5] = '\0';
 
 strncpy（）不会自动向dest字符串添加尾部'\ 0'。 如果源字符串长度大于len（strncpy的第三个参数），则len字符复制到dest而不尾随'\ 0'。 所以你必须通过代码显式地给出一个尾随'\ 0'。 

I think your output should be "some random string", because your two lines of code do nothing, see the comment. 
int main(){
   char token[] = "some random string";
   char c[23];  
   strcpy( c, token);
   strncpy(c, token, 5);  // Does nothing
   c[strlen(c)] = '\0';   // Does nothing
   cout<<c;
   return 0 ;
}
 
If you want to output "some", you could do this: 
strncpy(c, token, 5);
c[5] = '\0';
 
strncpy() doesn't automatically add trailing '\0' to dest string. If the source string length is bigger than len (3rd argument of strncpy), then len characters copied to dest without trailing '\0'. So you have to give a trailing '\0' explicitly by code.

Virtualenv不适用于RedHat(Virtualenv doesn't work on RedHat)

最满意答案

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