C++ unsigned int进行双重转换

Question

什么是正确的转换unsigned int方式double？我需要这是用于QCPCustomPlot数据设置,因为它QVector<double> 恰好作为创建图形的参数.

编辑:傻我."内存泄漏"错误问题是由于我QVector<double> x(time), y(ipv4int)错误地初始化而引起的.改变后的time值相同ipv4int(数据occurence频率)和x与y正确的匹配.

完成time变量,现在是有关转换的实际问题.如何将其转换为double包含值的格式,1855919686而不是格式为1.85592e+09？

QCustomPlot需要double但似乎QVector<double>不能像1.85592e + 09那样接受价值

更新的代码:

QVector<double> x(i), y(totalIP); //i=236052
for(int o = 0; o <= i; o++){
    double dSec = arrayIPxTime[o][0] - startSecond; //arrayIPTime[o][0] holds time in second
    double dMin = dSec/60;
    double ipv4addr = arrayIPxTime[o][1]; //arrayIPTime[o][0] holds ipaddr in integer format
    x[o] = dMin;
    //y[o] = ipv4addr; this is the line that causes crash. 
    qDebug()<<"Count "<<o<<" time "<< x[o] <<" ipv4 "<<ipv4addr<<" arrayIPxTime[o][1] "<<arrayIPxTime[o][1];
}

当前输出:

Count  236048  time  62.3167  ipv4  1.85592e+09  arrayIPxTime[o][1]  1855919686
Count  236049  time  62.3167  ipv4  1.85592e+09  arrayIPxTime[o][1]  1855919686
Count  236050  time  62.3167  ipv4  1.85592e+09  arrayIPxTime[o][1]  1855919686
Count  236051  time  62.3167  ipv4  1.85592e+09  arrayIPxTime[o][1]  1855919686

Answer 1

请放心,在转换unsigned int到double没有不导致内存泄漏.

写作double b = a;绰绰有余,或者只是传递a给需要double作为参数的函数.

请注意,对于IEEE754双精度浮点类型,转换精确到unsigned int最高为53的2次幂.