小编Ser*_*Ch.的帖子

我需要在保留文件名的同时将一些文件复制到二进制目录。到目前为止，我得到的是：

filegroup(
    name = "resources",
    srcs = glob(["resources/*.*"]),
)

genrule(
    name = "copy_resources",
    srcs = ["//some/package:resources"],
    outs = [ ],
    cmd = "cp $(SRCS) $(@D)",
    local = 1,
    output_to_bindir = 1,
)

现在，我必须在其中指定文件名，outs但似乎无法弄清楚如何解析标签以获得实际的文件名。

我正在使用这样的项目结构:

root
  |--build.gradle
  |--settings.gradle
  |
  |--child1
  |    |--build.gradle
  |
  |--child2
       |--build.gradle

在两个子项目中,build.gradle包含以下内容:

apply plugin: 'android'

...

android {
    ...

    signingConfigs {
        release {
            storeFile file("release.keystore")
            storePassword "pass"
            keyAlias "alias"
            keyPassword "pass"
        }
    }

    buildTypes {
        release {
            signingConfig signingConfigs.release
        }
    }
    ...
}

如何将signingConfigs部分移动到根项目以减少代码重复？

I'm running CI builds in a very resource-constrained Docker container.

AAPT2 doesn't seem to recognize that it's running in a container and spawns multiple daemon processes which then killed due to OOM and build fails:

AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon #0: Unexpectedly exit.
AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon #6: Unexpectedly exit.
AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon #3: Unexpectedly exit.
AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon #4: Unexpectedly exit.
AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon #2: Unexpectedly exit.
AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon #7: Unexpectedly exit.
AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon #1: Unexpectedly exit.
AAPT2 aapt2-3.3.0-beta01-5013011-linux Daemon …

我刚刚制作了这个程序,要求输入5到10之间的数字,然后计算这里输入的数字的总和就是代码

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
    int a,i,c;
    cout << "Enter the number between 5 and 10" << endl;
    cin >> a;
    if (a < 5 || a > 10)
    {
        cout << "Wrong number" << endl;
        system("PAUSE");
        return 0;
    }
    for(i=1; i<=a; i++)
    {
        c=c+i;
    }
    cout << "The sum of the first " << a << " numbers are " << c << endl;
    system("PAUSE");
    return 0;
}

如果我输入数字5,它应该显示

前5个数字的总和是15

但它显示出来

前5个数字的总和是2293687

但是当我把c设为0时 …