Makefile compare string input - shell

Question

To understand better string variables in a Makefile, I have tried to do this example :
KEYWORD=Nothing
test:
$(call myFunc)
define myFunc
ifeq ($(KEYWORD), "Apple")
echo "You have found the key"
else
echo "Try again"
endif
endef
But when I'm typing
make test KEYWORD="Fork"
It prints errors
syntax error near unexpected token `Fork,'
`ifeq (Fork, "Apple")'
I have also tried :
Put Apple or 'Apple' in ifeq
Put a space or not after "," : ifeq ($(KEYWORD), "Apple")
Run command with KEYWORD=Fork
Did it using shell (if [ ${KEYWORD} -eq "Apple" ])
I'm running out of ideas because I don't understand how Makefille / Shell interpret the assignment KEYWORD="Fork"
Thanks

Answer 1

MadScientist identifies the problem. Perhaps the solution you're looking for is simply evaluating the conditional earlier. eg:
KEYWORD ?= NOTHING
...
ifeq ($(KEYWORD), Apple)
define myFunc
echo "You have found the key"
endef
else
define myFunc
echo "Try again"
endef
endif

Answer 2

First, $(call myfunc) is 100% identical to writing $(myfunc). The call function in make merely expands a variable with some other local values (the arguments) bound first. If you don't provide any local values, then you're just expanding the variable.
Expanding a variable just replaces the variable reference with what it expands to. So writing:
FOO = bar
foo:
echo $(FOO)
is 100% identical to writing:
foo:
echo bar
So in your situation,
test:
$(call myFunc)
is the same as:
test:
$(myFunc)
which is the same as:
test:
ifeq ($(KEYWORD), "Apple")
echo "You have found the key"
else
echo "Try again"
endif
which is why you get the output you did: these are not valid shell commands, but since you've expanded the variable as part of a recipe, they are sent to the shell as part of the recipe.