[Bug] CompletableFuture 异步调用未录制

[abstinencetang]

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AREX Test Service

AREX Java Agent (arextest/arex-agent-java)

Current Behavior

dubbo服务作为入口，方法里使用CompletableFuture进行了两次异步调用，第二次调用依赖第一次返回结果，发现只录制到了第一次的dubbo请求，第二次没有录上。

我们async接口内部实现是调用的dubbo方法，这里为了简单复现问题，实现了两个service做查询，配置动态类模拟录制过程。

controller类：
@GetMapping("test/test12")
public String test12() {
CompletableFuture future = async(true, "abc").thenApply(result -> {
System.out.println("1getValue:" + result);
return result + "-线程1";
}).thenCompose(this::getFutureValue);
String s = "";
try {
s = future.get();
System.out.println("result:" + s);
} catch (Exception e) {
throw new RuntimeException(e);
}
return "success";
}

public CompletableFuture<String> async(boolean isConfig, String txt) {
    return CompletableFuture.supplyAsync(() -> {
        if (isConfig) {
            return ConfigService.getConfig(txt);
        } else {
            return CacheService.getValue(txt);
        }
    }, this.executorService);
}

private CompletableFuture<String> getFutureValue(String result) {
    return async(false, result).thenApply(re -> {
        System.out.println("2getValue:" + re);
        return re + "-线程2";
    });
}

service1：
public class ConfigService {

private static final ConfigService configService = new ConfigService();

public static ConfigService getInstance() {
    return configService;
}

private ConfigService() {
}

public static String getConfig(String key) {
    return "config-" + key;
}

}

service2：
public class CacheService {

private static final CacheService cacheService = new CacheService();

public static CacheService getInstance() {
    return cacheService;
}

private CacheService() {
}

public static String getValue(String key) {
    return key + "-" + key;
}

}

Expected Behavior

CompletableFuture 两次异步调用都能录制

Steps To Reproduce

""

Anything else

No response

Are you willing to submit a pull request to fix on your own?

• Yes I am willing to submit a pull request on my own!

[lucas-myx]

你好 感谢反馈，thenCompose本身就是异步线程创建的，内部又嵌套了一层async(false, result)异步方法，是否可以平铺开，怀疑是嵌套回调无法识别到线程的切换