问题描述
我给出了关于该函数可能执行的功能的大量指令，但是我对如何执行1的（1）和（2）感到困惑。请我将位置2设置在当前位置的左边。我不太了解是否应该创建新函数来实现此目标，或者是否可以使用已经完成的setBit（）函数？而且，如果确实需要创建新函数，我仍然对如何设置位以及回圈指令的含义感到困惑。
如果我能理解第一组指令并习惯于弄清楚如何设置位位置，我会更轻松地完成其余指令。我知道这是一个非常棘手的问题，但是我很迷失，不胜感激。如果有任何帮助或提示，或者朝正确的方向推进，我将不胜感激。
注意：只有主要功能才给我。我写了getBit，setBit和clearBit函数，但我觉得它们都不适用于问题1）。
我正在苦苦挣扎的说明：
(1) process the counter value,using the key value,as follows:
(a) make a copy of the counter value into a temp counter
(b) for every bit position,starting at bit position 7:
(i) compute two bit positions (position 1 and position 2) that you will use to perform an xor
between two of the temp counter bits: position 1 is set to the current bit position,and
position 2 is computed as follows:
(1) if the key bit at the current bit position is 1,then position 2 is set to one bit
position to the left of the current bit position,assuming we circle back to the
least significant bits (for example,we consider bit 0 to be to the left of bit 7)
(2) if the key bit at the current bit position is 0,then position 2 is set to two bit
positions to the left of the current bit position,assuming we circle back
(ii) xor the two temp counter bits found at positions 1 and 2
main.c
int main()
{
char str[8];
int choice;
printf("\nYou may:\n");
printf(" (1) Encrypt a message \n");
printf(" (2) Decrypt a message \n");
printf("\n what is your selection: ");
fgets(str,sizeof(str),stdin);
sscanf(str,"%d",&choice);
switch (choice) {
case 1:
break;
case 2:
break;
default:
break;
}
return 0;
}
unsigned char swapBitsOne(unsigned char counter,int bitOne,int bitTwo){
/* if the key bit at the current bit position is 1,then position 2 is set to one bit position to
the left of the current bit position,assuming we circle back to the least significant bits (for
example,we consider bit 0 to be to the left of bit 7)
*/
}
unsigned char swapBitsTwo(unsigned char counter,int bitTwo){
/*
if the key bit at the current bit position is 0,then position 2 is set to two bit positions to
the left of the current bit position,assuming we circle back
*/
}
unsigned char processCounter(unsigned char key,unsigned char counter){
unsigned char tempCounter;
strcpy(counter,tempCounter);
int i;
for(i = 7; i >= 0; --i){
if(key == 1){ // if the key bit at the current bit position is 1
counter = setBit(i,1
}
if(key == 0){
counter = swapBitsTwo();
}
}
return tempCounter;
}
unsigned char getBit(unsigned char c,int n){
return(c & 1 << n) >> n;
}
unsigned char setBit(unsigned char c,int n){
return c | (1 << n);
}
unsigned char clearBit(unsigned char c,int n){
return c & (~(1 << n));
}
解决方法
您有太多功能，使解决方案有点复杂。
首先，您不能在简单的strcpy参数上使用char。您需要传递char *参数。但是，只需使用简单的分配即可。
在processCounter中...
您需要从key中提取给定/当前位位置的位（例如unsigned char keybit = getBit(key,i);）。
“位置1”始终是当前位置（即i）。
“位置2”基于keybit是什么。
您需要根据c和pos1的两个值，从pos2获取给定的位。
您需要根据这两位的XOR值来设置/清除输出中的结果位。
无论如何，这里有一些重构的代码可以满足您的需求。
请注意，我已经颠倒了c和key参数的顺序，以使其更加惯用（并且与其他[helper]函数的c参数一致）。 / p>
此外，您实际上不需要“复制”参数c，只需将其他变量用作返回值即可。那是因为结果的所有八位都被替换了。
unsigned char
getBit(unsigned char c,int n)
{
#if ORIG
return (c & 1 << n) >> n;
#else
return (c >> n) & 1;
#endif
}
unsigned char
setBit(unsigned char c,int n)
{
return c | (1 << n);
}
unsigned char
clearBit(unsigned char c,int n)
{
#if ORIG
return c & (~(1 << n));
#else
return c & ~(1 << n);
#endif
}
unsigned char
processCounter(unsigned char c,unsigned char key)
{
int curpos;
int pos1;
int pos2;
unsigned char bit1;
unsigned char bit2;
unsigned char xor;
unsigned char out = 0;
for (curpos = 7; curpos >= 0; --curpos) {
// 1.b.i
pos1 = curpos;
// 1.b.i.1
if (getBit(key,curpos))
pos2 = (curpos + 1) % 8;
// 1.b.i.2
else
pos2 = (curpos + 2) % 8;
// 1.b.ii
bit1 = getBit(c,pos1);
bit2 = getBit(c,pos2);
xor = (bit1 ^ bit2) & 1;
if (xor)
out = setBit(out,curpos);
else
out = clearBit(out,curpos);
}
return out;
}
请注意，您可以将最终的if/else替换为：
out |= xor << curpos;
而且，您可以将1.b.i.1和1.b.i.2的if/else替换为：
pos2 = ((curpos + 2) - getBit(key,curpos)) % 8;