c语言long int转double 将无符号long long转换为C中的double

我意识到这个问题可能取决于处理器,但希望有人可以指出我正确的方向.对于我的生活,我无法弄清楚如何将表示纳秒的无符号long long int转换为表示C中的秒的double(我使用32位big-endian PowerPC 405进行此特定测试,以及GNU C99编译器).

我试过了:

unsigned long long int nanoseconds = 1234567890LLU;

double nanoseconds_d = nanoseconds*1e-9;

也:

nanoseconds_d = ((double) nanoseconds)*1e-9;

对于这两种情况,我只得到0.我在这里做错了什么？

编辑添加完整的例子

#include

#include

int

main( int argc, char *argv[] )

{

unsigned long long int nanoseconds = 1234567890LLU;

double nanoseconds_d = nanoseconds * 1e-9;

printf("%g\n", nanoseconds_d);

return 0;

}

MAKEFILE

SRCS = simple.c

INCLUDE := -I$(PWD)

CFLAGS := -O0 -g3 -Wall -fmessage-length=0 -mhard-float -fsigned-char -D_REENTRANT

LIBS := -lc

OBJS = $(SRCS:.c=.o)

PROG = $(SRCS:.c=).out

all: $(PROG)

$(PROG): $(OBJS)

@echo "Linking object files with output."

$(CC) -o $(PROG) $(OBJS) $(LIBS)

@echo "Linking complete."

$(OBJS): $(SRCS)

@echo "Starting compilation."

$(CC) $(CFLAGS) $(INCLUDE) -c $<

@echo "Compilation complete."

clean::

@$(RM) *.o *.out