本文整理汇总了C++中AstNodeModule::user2方法的典型用法代码示例。如果您正苦于以下问题：C++ AstNodeModule::user2方法的具体用法？C++ AstNodeModule::user2怎么用？C++ AstNodeModule::user2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AstNodeModule的用法示例。
在下文中一共展示了AstNodeModule::user2方法的2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: cantInline
void cantInline(const char* reason, bool hard) {
if (hard) {
if (m_modp->user2() != CIL_NOTHARD) {
UINFO(4," No inline hard: "<<reason<<" "<<m_modp<<endl);
m_modp->user2(CIL_NOTHARD);
m_statUnsup++;
}
} else {
if (m_modp->user2() == CIL_MAYBE) {
UINFO(4," No inline soft: "<<reason<<" "<<m_modp<<endl);
m_modp->user2(CIL_NOTSOFT);
}
}
}示例2: visit
// VISITORS
virtual void visit(AstNodeModule* nodep, AstNUser*) {
m_stmtCnt = 0;
m_modp = nodep;
m_modp->user2(CIL_MAYBE);
if (m_modp->castIface()) {
// Inlining an interface means we no longer have a cell handle to resolve to.
// If inlining moves post-scope this can perhaps be relaxed.
cantInline("modIface",true);
}
if (m_modp->modPublic()) cantInline("modPublic",false);
//
nodep->iterateChildren(*this);
//
bool userinline = nodep->user1();
int allowed = nodep->user2();
int refs = nodep->user3();
// Should we automatically inline this module?
// inlineMult = 2000 by default. If a mod*#instances is < this # nodes, can inline it
bool doit = ((allowed == CIL_NOTSOFT || allowed == CIL_MAYBE)
&& (userinline
|| ((allowed == CIL_MAYBE)
&& (refs==1
|| m_stmtCnt < INLINE_MODS_SMALLER
|| v3Global.opt.inlineMult() < 1
|| refs*m_stmtCnt < v3Global.opt.inlineMult()))));
// Packages aren't really "under" anything so they confuse this algorithm
if (nodep->castPackage()) doit = false;
UINFO(4, " Inline="<<doit<<" Possible="<<allowed<<" Usr="<<userinline<<" Refs="<<refs<<" Stmts="<<m_stmtCnt
<<" "<<nodep<<endl);
nodep->user1(doit);
m_modp = NULL;
}