Can't assign variable inside recipe

问题

How do I make this work? It errors out with "make: somevariable: Command not found"

sometarget:
    somevariable = somevalue

Full example:

CXXFLAGS = -I/usr/include/test -shared -fPIC

OBJ = main.o Server.o

blabla : $(OBJ) 
ifeq ($(argsexec),true) 
    # Creates an executable
    CXXFLAGS = -I/usr/include/test
    $(CXX) -o blabla $(OBJ) $(CXXFLAGS) 
else 
    # Creates a library
    DESTDIR = /home/pc
    $(CXX) -o blabla $(OBJ) $(CXXFLAGS) 
    ./bn.sh
endif

回答1:

I found a solution using the eval function:

$(eval variablename=whatever)

This works :)

(although I may now try to find an easier build system ;))

Thanks everyone for reading and also of course @eriktous for writing!

回答2:

If you write it like you did, the assignment will be executed as a shell command, which gives the error you got.

I would try organising it something like this:

CXXFLAGS = -I/usr/include/test
ifneq ($(argsexec),true) 
  CXXFLAGS += -shared -fPIC
  DESTDIR = /home/pc
endif

OBJ = main.o Server.o

blabla : $(OBJ) 
    $(CXX) -o blabla $(OBJ) $(CXXFLAGS) 
ifneq ($(argsexec),true) 
    ./bn.sh
endif

This should do what you want, although I'm not quite happy with using the ifneq construct twice. I'd have to think harder to come up with something that avoids that.

来源：https://stackoverflow.com/questions/6519234/cant-assign-variable-inside-recipe