以下c ++代码返回什么?

smt*_*tsp -1 c++ types iterator unordered-map map

我有一段代码需要理解.但是我迷失了一点.这是代码:

typedef unordered_map <string, TimeStampSet *> HIEMap;
typedef set <TimeStamp> TimeStampSet;

struct HostInfo {
    HostActivity *hostActivity;
    HIEMap *hieMapArr;
};

typedef unordered_map <uint32_t, HostInfo *> HostInfoMap;

HIEMap::iterator hieMapIt;

void method(...){
    for (hieMapIt = hostInfoIt -> second -> hieMapArr -> begin();
         hieMapIt != hostInfoIt -> second -> hieMapArr -> end();
         hieMapIt = nextMapIt)
    {
        if (hieMapIt -> second == NULL) {
           //what does *hieMapIt -> second* returns?
        }
    }
}
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什么hieMapIt -> second回报?我有点迷茫.

这不是所有代码,都有初始化等等.但是我没有把所有的代码放在这里.

谢谢,

Bar*_*art 5

它会返回一个TimeStampSet *.也就是说,unordered_map的第二部分(字符串,TimeStampSet*)"对".