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問(wèn)題描述
我很好奇，如何總結(jié)java8流中的多個(gè)變量.
I am curious, how to sum up multiple variables in a java8 stream.
Integer wCPU = 0;
Double wnetwork = 0.0;
Double wMem = 0.0;
this.slaContractList.forEach(sla -> {
wCPU += sla.getNumberOfCPUs();
wnetwork += sla.getNetworkBandwith();
wMem += sla.getMemory();
});
但是，這不會(huì)編譯，因?yàn)?lambda 表達(dá)式中的變量應(yīng)該是 final.
However, this does not compile as the variable in the lambda expression should be final.
推薦答案
假設(shè) slaContractList 是 SlaContract 對(duì)象的列表，它有構(gòu)造函數(shù) SlaContract(numberOfCPUs, networkBandwith, memory) 你可以:
Assuming slaContractList is a list of SlaContract objects, and it has constructor SlaContract(numberOfCPUs, networkBandwith, memory) you can:
SlaContract sumContract = slaContractList.stream()
.reduce(new SlaContract(0, 0.0, 0.0), (sla1, sla2) -> {
return new SlaContract(sla1.getNumberOfCPUs() + sla2.getNumberOfCPUs(), sla1.getworkBandwith() + sla2.getworkBandwith(), sla1.getMemory() + sla2.getMemory());
});
Double wnetwork = sumContract.getworkBandwith();
Double wMem = sumContract.getMemory();
Integer wCPU = sumContract.getNumberOfCPUs();
相同的解決方案，但對(duì)于簡(jiǎn)單的類:
The same solution, but for simple class:
Point sumPoint = pointsList.stream()
.reduce(new Point(0, 0), (p1, p2) -> {
return new Point(p1.x + p2.x, p1.y + p2.y);
});
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