无法在配方内分配变量 [英] Can't assign variable inside recipe

问题描述

我如何进行这项工作?错误显示为"make:somevariable:找不到命令"

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

sometarget:
    somevariable = somevalue

完整示例:

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

推荐答案

我找到了使用

I found a solution using the eval function:

$(eval variablename=whatever)

这有效:)

(尽管我现在可以尝试找到一个更简单的构建系统；))

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

感谢大家的阅读，当然也感谢@eriktous的写作！

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

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