将无符号long long转换为C中的double

Question

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 <stdio.h>
#include <stdlib.h>

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

Answer 1

%g使用打印时在此处起作用

#include <stdlib.h>
#include <stdio.h>

int main(){
    unsigned long long int nanoseconds = 1234567890LLU;
    double nanoseconds_d = nanoseconds*1e-9;
    printf("%g\n", nanoseconds_d);

    nanoseconds_d = ((double) nanoseconds)*1e-9;
    printf("%g\n", nanoseconds_d);
}

输出

1.23457
1.23457