C无符号int指针之间的XOR如何

Question

我有两个带32位的无符号int指针，我想在这些无符号指针之间进行XOR运算。

char* a = "01110011011100100110111101000011";
char* b = "10111001100011001010010110111101";
unsigned* au = (unsigned*) a;
unsigned* bu = (unsigned*) b;
unsigned* cu = a ^ b;

错误是：

 invalid operands to binary ^ (have ‘unsigned int *’ and ‘unsigned int *’)

Answer 1

你有字符串，而不是无符号整数。在对它们执行按位操作之前，您需要将它们转换为无符号整数：

char* a = "01110011011100100110111101000011";
char* b = "10111001100011001010010110111101";
unsigned au = strtoul(a, 0, 2);
unsigned bu = strtoul(b, 0, 2);
unsigned cu = a ^ b;

Answer 2

你的代码需要更加努力才能解决这个问题。

1. 迭代字符串。
2. 从数字中拉出数字。
3. 在数字之间执行XOR。
4. 将数字转换为数字。
5. 将结果数字添加到输出中。

char str1[] = "01110011011100100110111101000011";
char str2[] = "10111001100011001010010110111101";
char* p1 = str1;
char* p2 = str2;
for ( ; *p1 != '\0' && *p2 != '\0'; ++p1, ++p2 )
{
   unsigned int n1 = *p1 - '0';
   unsigned int n2 = *p2 - '0';
   unsigned int n = n1 ^ n2;
   char c = n + '0';

   // Now you can change either a or be to contain the output.
   *p1 = c;
}

// At this point, str1 should contain a string that looks like you performed an XOR between str1 and str2.