node.js configure file syntax error line 433
node.js
Solution
The issue was that the python script attempting to run the configure file was Python 2.4. So I installed Python 2.7 (I may have had it already), and I ran the script again using:
python2.7 configure
Then it ran correctly.
Problem
I downloaded node.js through a link that was not joyent, and it gave me an old version of node. So I wanted to reinstall node.js with the new git://github.com/joyent/node.git. So I tried running the same clone script over and it gave a fatal error because my node folder isn't empty. So I deleted the node folder and ran it again and it cloned it. After that I tried running the configure file and I got an error: ``` File "./configure", line 433 fpu = 'vfpv3' if armv7 else 'vfpv2' ^ SyntaxError: invalid syntax ``` I have no idea how to solve a problem? I've tried deleting the node folder and cloning the git again but same error. Here is the code I found in the file near line 433: ``` armv7 = is_arch_armv7() # CHECKME VFPv3 implies ARMv7+ but is the reverse true as well? fpu = 'vfpv3' if armv7 else 'vfpv2' ```