bash: cannot execute binary file

Question

I am trying to get compile some C code from within Cygwin on 64-bit Windows and call the executable using some parameters. The compilation goes through fine (the Makefile is below). But when I run the executable, I get the error cannot execute binary file.

A quick shows that indeed the binary is built for Linux.

ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.24, BuildID[sha1]=0x7d39e30cb249d2047815de23a9df33fccf91a58f, not stripped

How can I change the Makefile so that it compiles for Windows?

Here is the Makefile

.SUFFIXES: .c .u
CC= gcc

CFLAGS = -g -Wall -O3 -DHAVE_INLINE -DGSL_RANGE_CHECK_OFF
# LDFLAGS = -lm -lgsl -latlas -lgslcblas
LDFLAGS = -lm -lgsl -lgslcblas
# LDFLAGS = -lm -lgsl -latlas -lcblas
LOBJECTS= a.o b.o c.o 
LSOURCE= a.c b.c c.c 

windows: $(LOBJECTS)
    $(CC) $(LOBJECTS) -o stm $(LDFLAGS)

clean:
    -rm -f *.o

jim mcnamara · Accepted Answer · 2013-06-07 18:58:04Z

4

cygwin is 32 bits, not 64 bits. This is the error you get when trying to run a 64 bit program in a 32 bit environment.

edit: Oh, and you can't compile C for Windows from inside cygwin. There are C compilers for Windows, I know little about them.

answered Jun 7, 2013 at 18:58

jim mcnamara

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1 Comment

ollo Over a year ago

Or the MinGW fork MinGW-w64, which also has 64bit support.

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