09秋数据库课程设计_小结_ 5

今天真是可喜可贺了，课程设计的第二个检查点，今天检查代码量了。嗯，其实呢，到导师检查的时候完成量大致50%，可是今天课时结束后，我们代码的工作基本完成了，可以说是形成了demo了，看来超量完成任务是没问题啦~~~~~~~~~~

其实，今天完成的代码量不在少数，收获颇丰啊，总结一下~~~~~~

1、对“warning: missing braces around initializer”的处理：

在进行变换时，需要用到S变换表，而我使用数组来存储的：

char *arrnS1[][16] = {"1110","0100","1101","0001","0010","1111","1011","1000","0011","1010","0110","1100","0101","1001","0000","0111","0000","1111","0111","0100","1110","0010","1101","0001","1010","0110","1100","1011","1001","0101","0011","1000","0100","0001","1110","1000","1101","0110","0010","1101","1111","1100","1001","0111","0011","1010","0101","0000","1111","1100","1000","0010","0100","1001","0001","0111","0101","1101","0011","1110","1010","0000","0110","1101"};

char *arrnS1[][16] = {{"1110","0100","1101","0001","0010","1111","1011","1000","0011","1010","0110","1100","0101","1001","0000","0111"},"0000","1111","0111","0100","1110","0010","1101","0001","1010","0110","1100","1011","1001","0101","0011","1000"},"0100","0001","1110","1000","1101","0110","0010","1101","1111","1100","1001","0111","0011","1010","0101","0000"},"1111","1100","1000","0010","0100","1001","0001","0111","0101","1101","0011","1110","1010","0000","0110","1101"}};

上面的初始化方法就会引起该warning，下面那个就不会。看出区别了吗？实际上，我初始化了一个4行16列的二维数组，区别在于，下面那个在初始化时，对每行（16列）都加上了“{}”括起来，这样表示它们是一行，就不会出现warning了。

结论：对于初始化，需要严格按照定义，对于结构中的数组，必须加{}。

2、参数传递的问题

同组的伙伴有一部分是这样做的:

out=rewindReplace(out,3);

tmpPrintf=out;
printf("here is the after rewindReplace 3:\n");
for (int counter=0;counter<64;counter++)
{
printf("%d ",*tmpPrintf);
tmpPrintf++;
j=counter+1;
j%=8;
if (j==0)
{
printf("\n");
}
}
printf("%d\n",out);
printf("%d\n",*out);

out=rewindProductChange(out);

tmpPrintf=out;
printf("here is the after rewindProductChange :\n");
for (int counter=0;counter<64;counter++)
{
printf("%d ",*tmpPrintf);
tmpPrintf++;
j=counter+1;
j%=8;
if (j==0)
{
printf("\n");
}
}

在两次输出out的值时，就会出错，而前面的输出完全正确，但是在“rewindProductChange”里又没有将out的值进行改变，为什么两次的输出不同呢？

经过思考，我们发现在“rewindProductChange”中，定义了形参“int *”，而并没有对其进行“new”，这样，在函数尾部虽然return了，但是随着函数的结束，该变量的内存空间也被释放（由于其内存空间跟函数的内存空间在一起）。我们的解决办法是，对其进行“new”，这样它的内存空间就独立于函数本身，即使函数退出，其内存空间也不会被释放。