Question

看着|，它被描述为按位运算符OR。因此，在此代码示例中：

private int getColorRGB(int color) { // 255255255 would be white, 000255000 green, etc.

    if (color < 0) return -1;

    int r = color / 1000000 % 1000;
    int g = color / 1000 % 1000;
    int b = color % 1000;

    if (r > 255 || g > 255 || b > 255) throw new IllegalArgumentException("RGB values cannot exceed 255.");

    return (r >> 16) | (g >> 8) | b; // POINT OF INTEREST
}

我可以用|将标记为POINT OF INTEREST的行中的2 +替换为rrrgggbbb，

该方法采用整数200200200，因此255为蓝色，function Update-program { Clear-Host Write-Host "Updating Program..." Write-Host " " if ( -not (Test-Path "$downloadLocation\Here" -PathType Container)) { New-Item -Path "$downloadLocation\Here" -ItemType directory -Force } $apiUrl = 'https://ci.appveyor.com/api' $headers = @{ "Authorization" = "Bearer $token" "Content-type" = "application/json" } $accountName = 'Hello' $projectSlug = 'World' # get project with last build details $project = Invoke-RestMethod -Method Get -Uri "$apiUrl/projects/$accountName/$projectSlug" -Headers $headers # we assume here that build has a single job # get this job id $jobId = $project.build.jobs[0].jobId # get job artifacts (just to see what we've got) $artifacts = Invoke-RestMethod -Method Get -Uri "$apiUrl/buildjobs/$jobId/artifacts" -Headers $headers # here we just take the first artifact, but you could specify its file name # $artifactFileName = 'MyWebApp.zip' $artifactFileName = $artifacts[0].fileName # artifact will be downloaded as $localArtifactPath = "$downloadLocation\$artifactFileName" # download artifact # -OutFile - is local file name where artifact will be downloaded into # the Headers in this call should only contain the bearer token, and no Content-type, otherwise it will fail! Invoke-RestMethod -Method Get -Uri "$apiUrl/buildjobs/$jobId/artifacts/$artifactFileName" ` -OutFile $localArtifactPath -Headers @{ "Authorization" = "Bearer $token" } Clear-Host Write-Host "Download complete, extracting..." Write-Host " " Expand-Archive -LiteralPath "$downloadLocation\$artifactFileName" -DestinationPath "$ProgramDir" -Force Clear-Host Write-Host "Program updated" Write-Host " "为浅灰色，等等。

所以，我的问题是；两者有什么区别

Answer 1

a = 2; // binary 0x10
b = 2; // binary 0x10

c = a + b; // c = 4
c = a | b; // c = 2

|是位操作，不等于+

有时它会给出相同的结果：例如2+1和2|1；但这不是规则

Answer 2

|进行按位或运算，即取两个数字中较大的对应位，而+进行两个对应位的相加，并取进一步的进位，表示1+1给出10而1|1只会停止到1。如果较大的参数在范围内，则|永远不会导致超出范围，而如果两个参数之和导致数字超出范围，则+可能会超出范围。如果两个数字中的对应位不同，则只有在这种情况下，|才充当+，因为在求和期间，进位永不产生。

Answer 3

记住|和+是2个不同的运算符。话虽如此，有时它们可以具有相同的结果，例如1 * 1和1/1。虽然它们具有相同的结果，但它们不会经过相同的过程。

按位运算符|和+一样？

3 个答案: