本文整理汇总了C++中IOdictionary::instance方法的典型用法代码示例。如果您正苦于以下问题：C++ IOdictionary::instance方法的具体用法？C++ IOdictionary::instance怎么用？C++ IOdictionary::instance使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类IOdictionary的用法示例。
在下文中一共展示了IOdictionary::instance方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: main
int main(int argc, char *argv[])
{
argList::noParallel();
argList::addBoolOption("rewrite");
argList::addBoolOption("show");
argList args(argc, argv);
Time runTime(args.rootPath(), args.caseName());
const word dictName("fvSolution");
bool optRewrite = args.optionFound("rewrite");
bool optShow = args.optionFound("show");
IOdictionary solutionDict
(
IOobject
(
dictName,
"system",
runTime,
IOobject::MUST_READ_IF_MODIFIED,
IOobject::NO_WRITE,
false
)
);
if (!solutionDict.found("solvers"))
{
Info<<"no solvers entry found in : " << dictName << endl;
return 2;
}
if (optRewrite && solutionDict.instance() != "system")
{
Info<<"instance is not 'system' "
"- disabling rewrite for this file" << nl;
optRewrite = false;
}
dictionary& solverDict = solutionDict.subDict("solvers");
wordList names = solverDict.toc();
wordList oldNames = names;
bool changed = false;
for (label orig = 0; orig < names.size()-1; ++orig)
{
// skip patterns or entries that have already been done
if (names[orig].empty() || wordRe::isPattern(names[orig]))
{
continue;
}
const dictionary& dict1 = solverDict.subDict(names[orig]);
for (label check = orig+1; check < names.size(); ++check)
{
// skip patterns or entries that have already been done
if (names[check].empty() || wordRe::isPattern(names[check]))
{
continue;
}
const dictionary& dict2 = solverDict.subDict(names[check]);
// check for identical content
if (checkDictionaryContent(dict1, dict2))
{
names[orig] += "|" + names[check];
names[check].clear();
changed = true;
}
}
}
if (changed)
{
forAll(names, nameI)
{
if (names[nameI].empty())
{
solverDict.remove(oldNames[nameI]);
Info<<" #remove " << oldNames[nameI];
}
else
{
Info<< " " << oldNames[nameI];
if (names[nameI] != oldNames[nameI])
{
// make "(abc|def)" pattern
keyType renamed( "(" + names[nameI] + ")", true);
solverDict.changeKeyword(oldNames[nameI], renamed);
Info<< " -> " << renamed;
}
}
//.........这里部分代码省略.........