Debugger: use QString::fromLatin1 instead of QString::fromAscii

By source - latin1 is really expected as there is no any check
or usage of QTextCodec::setCodecForCStrings() currently.

QString::fromAscii() might break 'Latin1' input in some cases.

A quote from documentation about QString::fromAscii():

"Note that, despite the name, this function actually uses the
codec defined by QTextCodec::setCodecForCStrings() to convert str
to Unicode. Depending on the codec, it may not accept valid
US-ASCII (ANSI X3.4-1986) input. If no codec has been set, this
function does the same as fromLatin1()."

Change-Id: I49cf047ca674d2ec621b517c635d1927bb2e796f
Reviewed-by: Friedemann Kleint <[email protected]>
Reviewed-by: hjk <[email protected]>

1 files changed, 3 insertions, 3 deletions

diff --git a/src/plugins/debugger/watchhandler.cpp b/src/plugins/debugger/watchhandler.cpp
index 5215cf8c47d..79264f95d40 100644
--- a/src/plugins/debugger/watchhandler.cpp
+++ b/src/plugins/debugger/watchhandler.cpp
@@ -713,14 +713,14 @@ int WatchModel::itemFormat(const WatchData &data) const
 static inline QString expression(const WatchItem *item)
 {
     if (!item->exp.isEmpty())
-         return QString::fromAscii(item->exp);
+         return QString::fromLatin1(item->exp);
     if (item->address && !item->type.isEmpty()) {
-        return QString::fromAscii("*(%1*)%2").
+        return QString::fromLatin1("*(%1*)%2").
                 arg(QLatin1String(item->type), QLatin1String(item->hexAddress()));
     }
     if (const WatchItem *parent = item->parent) {
         if (!parent->exp.isEmpty())
-           return QString::fromAscii("(%1).%2")
+           return QString::fromLatin1("(%1).%2")
             .arg(QString::fromLatin1(parent->exp), item->name);
     }
     return QString();