gfortran: Why does this fail to compile?
Mark Hounschell
markh@compro.net
Tue Jan 22 14:12:00 GMT 2013
Can someone tell me why this fails to compile in gfortran? I've used
this type of format specification in many different fortran compilers.
gfortran -v
Using built-in specs.
COLLECT_GCC=gfortran
COLLECT_LTO_WRAPPER=/usr/lib/gcc/i586-suse-linux/4.7/lto-wrapper
Target: i586-suse-linux
Configured with: ../configure --prefix=/usr --infodir=/usr/share/info
--mandir=/usr/share/man --libdir=/usr/lib --libexecdir=/usr/lib
--enable-languages=c,c++,objc,fortran,obj-c++,java,ada
--enable-checking=release --with-gxx-include-dir=/usr/include/c++/4.7
--enable-ssp --disable-libssp --disable-libitm --disable-plugin
--with-bugurl=http://bugs.opensuse.org/ --with-pkgversion='SUSE Linux'
--disable-libgcj --disable-libmudflap --with-slibdir=/lib
--with-system-zlib --enable-__cxa_atexit
--enable-libstdcxx-allocator=new --disable-libstdcxx-pch
--enable-version-specific-runtime-libs --enable-linker-build-id
--program-suffix=-4.7 --enable-linux-futex --without-system-libunwind
--with-arch-32=i586 --with-tune=generic --build=i586-suse-linux
Thread model: posix
gcc version 4.7.1 20120723 [gcc-4_7-branch revision 189773] (SUSE Linux)
program test_c3
IMPLICIT NONE
c
c The first read below, reads a line of ascii text
c containing 20 space separated, 4 char, hex fields
c into a buffer.
c The format of the file is
c 1 2 3 4 17 18 19 20
c 0002 0002 3130 0052 .... .... ff4b 0000 1024 b070
c
c The second read is to read the text fields into
c an array of integer*2s but the "valid" format
c specifier in that read is invalid to gfortran
c for some reason. Why????
c
INTEGER*4 WYPTUNIT/100/
INTEGER*2 WYPTBUFF(20)
CHARACTER*100 READBUFF
OPEN(UNIT = WYPTUNIT, FILE = 'waypthex.txt')
READ(WYPTUNIT, '(A100)') READBUFF
READ(READBUFF, '(19(Z4,X1),Z4)') WYPTBUFF
STOP
END
#gfortran test_c3.f
test_c3.f:23.37:
READ(READBUFF, '(19(Z4,X1),Z4)') WYPTBUFF
1
Error: Unexpected element ')' in format string at (1)
Thanks in advance
Mark
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