Makefile: remove accidental recipe prefix in conditional

Back in 728b9ac0c3 (Makefile(s): avoid recipe prefix in conditional
statements, 2024-04-08), we prepared our Makefiles for a forthcoming
change in upstream Make that would ban the recipe prefix within a
conditional statement by replacing tabs (the prefix) with eight spaces.

In b9d6f64393 (compat/zlib: allow use of zlib-ng as backend,
2025-01-28), a handful of recipe prefix characters were introduced in a
conditional statement ('ifdef ZLIB_NG'), causing 'make' to fail on my
system, which uses GNU Make 4.4.90.

Remove the recipe prefix characters by replacing them with the same
script as is mentioned in 728b9ac0c3.

Signed-off-by: Taylor Blau <me@ttaylorr.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

diff --git a/Makefile b/Makefile
index 1853e6ddfafbc243d449b62410c003d00f555e0e..82e51a80bd6a702d771cde76a48b24b738607aa3 100644 (file)
--- a/Makefile
+++ b/Makefile
@@ -1690,16 +1690,16 @@ IMAP_SEND_LDFLAGS += $(OPENSSL_LINK) $(OPENSSL_LIBSSL) $(LIB_4_CRYPTO)
 
 ifdef ZLIB_NG
        BASIC_CFLAGS += -DHAVE_ZLIB_NG
-       ifdef ZLIB_NG_PATH
+        ifdef ZLIB_NG_PATH
                BASIC_CFLAGS += -I$(ZLIB_NG_PATH)/include
                EXTLIBS += $(call libpath_template,$(ZLIB_NG_PATH)/$(lib))
-       endif
+        endif
        EXTLIBS += -lz-ng
 else
-       ifdef ZLIB_PATH
+        ifdef ZLIB_PATH
                BASIC_CFLAGS += -I$(ZLIB_PATH)/include
                EXTLIBS += $(call libpath_template,$(ZLIB_PATH)/$(lib))
-       endif
+        endif
        EXTLIBS += -lz
 endif