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现象：
QString x; qDebug() << __LINE__<< x.size() << x.length();//10 0 0 qDebug() <<
__LINE__<< x.[0];//11 '\x0' qDebug() << __LINE__<< x.[10];//12 '\x0' qDebug()
<< __LINE__<< x.at(0); //ASSERT: "uint(i) < uint(size())" in file
/usr/include/x86_64-linux-gnu/qt5/QtCore/qstring.h, line 931
QString()[k]，k为任意值都不会崩溃
QString().at(0)，则会崩溃
原因：
QString()[10] <=> QString().operator[](10);
源码：QString/operator[]
inline QCharRef QString::operator[](int i) { Q_ASSERT(i >= 0); return
QCharRef(*this, i); }
源码：QString/QCharRef
class Q_CORE_EXPORT QCharRef { QString &s; int i; inline
QCharRef(QString &str, int idx) : s(str),i(idx) {} friend class
QString; public: // most QChar operations repeated here // all this is
not documented: We just say "like QChar" and let it be. inline operator
QChar() const { return i < s.d->size ? s.d->data()[i] : 0; }
看出operator[]的函数，传任意值是不会崩溃的
备注
下图能看出来，QString的有
两个operator[]()：QString()[10]不会崩溃
两个operator[]() const：QString()[10]会崩溃
一个at() const;：QString().at(10)会崩溃
当我们外部调用operator[]函数时，优先operator[]()，其次operator[]() const
demo验证
class A{ public: A(){} int printx() const {qDebug() << "printx() const";} int
printx() {qDebug() << "printx() ";} }; int main(int argc, char *argv[]) {
QApplication a(argc, argv); A aa; aa.printx();//printx()