将Char*映射到两个整数-MappingaChar*totwoIntegers

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How about using only one set implemented by binary tree?

如何只使用二叉树实现的一组?

Key:
     {name, license, ssid}

Function Compare:

     name_compare(name1, name2) || license_compare(license1, license2) || ssid_compare(ssid1, ssid2)

This is an O(logN) search.

这是一个O(logN)搜索。

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It would be useful to give some detail on the values. If the values were originally integers, which have been converted to a character array for transport, than the problem can be solved by casting it to an int. Otherwise, your best option is to use stdlib.h functions like atoi() and atoc().

提供有关值的一些细节会很有用。如果这些值最初是整数,已经转换为用于传输的字符数组,那么问题可以通过将其转换为int来解决。否则,您最好的选择是使用像atoi()和atoc()这样的stdlib.h函数。

Based on your example code it looks like you are trying to extract two values from the same char array. If this is the case I would recommend using memcpy() like this.

根据您的示例代码,您似乎尝试从同一个char数组中提取两个值。如果是这种情况,我建议使用像这样的memcpy()。

/* Based on what you wrote above I am assuming name is 8 bytes  */
int firstValueSize = 4/* whatever size in bytes the first number should be */;
int secOndValueSize= 4/* Same with this */;
int *firstValue = (int)malloc(sizeof(int));
int *secOndValue= (int)malloc(sizeof(int));
memcpy(firstValue, name, sizeof(firstValueSize));
memcpy(secondValue, name + firstValueSize, sizeof(secondValueSize));

This should work if they were originally integers. Otherwise you will have to use the functions mentioned above.

如果它们最初是整数,则应该有效。否则你将不得不使用上面提到的功能。