waf가 선택한 도구를 결정하는 방법은 무엇입니까?

Question

'cxx_compiler'등으로 선택된 특정 도구 waf를 어떻게 확인할 수 있습니까? Exempli gratia에 :

def configure(ctx): 
    print('Running ' + ctx.cmd + ' in ' + ctx.path.abspath()) 
    ctx.load('compiler_c') 
    ctx.load('compiler_cxx') 

def build(ctx): 
    print('Running ' + ctx.cmd + ' in ' + ctx.path.abspath()) 
    # Here print which C++ compiler was chosen 
    print 'Building cpp files with %s' % WHAT_GOES_HERE 

Answer 1

def options(opt): 
    opt.load('compiler_c') 
    opt.load('compiler_cxx') 

def configure(cfg): 
    cfg.load('compiler_c') 
    cfg.load('compiler_cxx') 

def build(bld): 
    print "Compiler is CC_NAME %s CC %s"%(bld.env.CC_NAME,bld.env.CC) 
    print "Compiler is CXX_NAME %s CXX %s"%(bld.env.CXX_NAME,bld.env.CXX) 

당신에게 줄 것이다 :

D:\temp>waf.bat configure build --check-c-compiler=gcc --check-cxx-compiler=g++ 
Setting top to       : D:\temp 
Setting out to       : D:\temp\build 
Checking for 'gcc' (c compiler)   : c:\tools\gcc\bin\gcc.exe 
Checking for 'g++' (c++ compiler)  : c:\tools\gcc\bin\g++.exe 
'configure' finished successfully (0.191s) 
Waf: Entering directory `D:\temp\build' 
Compiler is CC_NAME gcc CC ['c:\\tools\\gcc\\bin\\gcc.exe'] 
Compiler is CXX_NAME gcc CXX ['c:\\tools\\gcc\\bin\\g++.exe'] 
Waf: Leaving directory `D:\temp\build' 
'build' finished successfully (0.008s)