make ignores -std=c99 flag when compiling and linking a C program

Question

I've tried to get my makefile to compile a file that requires -std=c99 to run. In this case, its to get a "for-loop" through.

This is my code, (its been used "tab" before "$(CC)" ):

CC = gcc
CFLAGS = -c -std=c99

...

Download.o : Download.c
    $(CC) $(CFLAGS) Download.c

Download.c contains methods used to download elements from the web

Error message

$ make
gcc -c -std=c99 Download.c
gcc Download.c -o Program
Download.c: In function ‘downloadImageparts’:
Download.c:11:2: error: ‘for’ loop initial declarations are only allowed in C99 mode
Download.c:11:2: note: use option -std=c99 or -std=gnu99 to compile your code
Download.c:13:3: error: ‘for’ loop initial declarations are only allowed in C99 mode
make: *** [comp] Error 1

Attemt to debug

If I run gcc -c -std=c99 Download.c in terminal it works fine.

This problems appears when run in Linux.

SOLVED:

I created a dummy project to show my lecturer, in an attempt to solve my problem. In the dummy project all works fine with the code described. For some reason my code works on place but not in the other. If someone reading this having the same problem as me and would like to see an example project. let me know and I'll write the code here. Thanks

Answer 1

You're looking at the wrong rule. Download.c is actually compiling fine, but the linking stage is wrong.

$ make
gcc -c -std=c99 Download.c  # Compile
gcc Download.c -o Program   # Link

Fix the make rule that links the program. It should probably look something like this:

Program: a.o b.o c.o
    $(CC) $(LDFLAGS) -o $@ $^ $(LIBS)

While you're at it, I suggest a more complete Makefile will look something like this:

all: Program
clean:
    rm -f Program *.o
.PHONY: all clean

# -c is implicit, you don't need it (it *shouldn't* be there)
# CC is also implicit, you don't need it
CFLAGS := -std=c99 -g -Wall -Wextra

Program: a.o b.o c.o
    $(CC) $(LDFLAGS) -o $@ $^ $(LIBS)

# Make will automatically generate the necessary commands
# You just need to name which headers each file depends on
# (You can make the dependencies automatic but this is simpler)
a.o: a.c header.h
b.o: b.c header.h header2.h
c.o: c.c header.h

Examples of how to do it wrong

Linker flags are actually fairly touchy! Be sure to type in the line above exactly as I have written it, and don't assume that what you've written is equivalent. Here are some examples of slightly different commands that are wrong and should not be used:

# WRONG: program must depend on *.o files, NOT *.c files
Program: a.c b.c c.c
    $(CC) ...

# WRONG: -c should not be in CFLAGS
CFLAGS := -c -std=c99

Program: a.o b.o c.o
    # WRONG: $(CFLAGS) should not be here
    # you are NOT compiling, so they do not belong here
    $(CC) $(CFLAGS) $(LDFLAGS) -o $@ $^ $(LIBS)

    # WRONG: $(LIBS) MUST come at the end
    # otherwise linker may fail to find symbols
    $(CC) $(LDFLAGS) -o $@ $(LIBS) $^

    # WRONG: do not list *.o files, use $^ instead
    # otherwise it is easy to get typos here
    $(CC) $(LDFLAGS) -o $@ a.o b.o c.o $(LIBS)

    # WRONG: $(LDFLAGS) must be at the beginning
    # it only applies to what comes after, so you
    # MUST put it at the beginning
    $(CC) -o $@ $(LDFLAGS) $^ $(LIBS)

    # WRONG: -c flag disables linking
    # but we are trying to link!
    $(CC) $(LDFLAGS) -c -o $@ $^ $(LIBS)

    # WRONG: use $(CC), not gcc
    # Don't sabotage your ability to "make CC=clang" or "make CC=gcc-4.7"
    gcc $(LDFLAGS) -o $@ $^ $(LIBS)

    # WRONG: ld does not include libc by default!
    ld $(LDFLAGS) -o $@ $^ $(LIBS)