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>

Taylor Blau committed Feb 13, 2025 at 15:25 UTC f23179924bf4ee0e888cfbe911d9bd472918bcb4
1 file changed +4 -4
Makefile
+4 -4
@@ -1690,16 +1690,16 @@ IMAP_SEND_LDFLAGS += $(OPENSSL_LINK) $(OPENSSL_LIBSSL) $(LIB_4_CRYPTO)
1690
1691 ifdef ZLIB_NG
1692 BASIC_CFLAGS += -DHAVE_ZLIB_NG
1693 - ifdef ZLIB_NG_PATH
1693 + ifdef ZLIB_NG_PATH
1694 BASIC_CFLAGS += -I$(ZLIB_NG_PATH)/include
1695 EXTLIBS += $(call libpath_template,$(ZLIB_NG_PATH)/$(lib))
1696 - endif
1696 + endif
1697 EXTLIBS += -lz-ng
1698 else
1699 - ifdef ZLIB_PATH
1699 + ifdef ZLIB_PATH
1700 BASIC_CFLAGS += -I$(ZLIB_PATH)/include
1701 EXTLIBS += $(call libpath_template,$(ZLIB_PATH)/$(lib))
1702 - endif
1702 + endif
1703 EXTLIBS += -lz
1704 endif
1705