为了在C ++中调试，如何声明一个不被优化的变量（放入寄存器）？

Question

I'm developing a simple application in C++/Qt , and I have the following declaration:

QGridLayout *layout = new QGridLayout;

I'm debugging the application using gdb . I set a breakpoint, it works fine, and the debugger hits the line. But if I try to inspect the object declared above I get this output:

 -data-evaluate-expression --thread 1 --frame 0 layout ^done,value="<value> optimized out>" 

I read that this message, "<value> optimized out>" , occurs because the compiler optimized the code, and put the data into the register. I'm using g++ compiler, with flag -O0 (none optimization) set.

Is there something I'm missing, or does it exist a way to declare a variable not to be optimized, say, as opposed to the storage specifier register ? I'm on Ubuntu 10.10 Maverick, kernel 2.6.35-24.

EDIT1

Some more code:

WorkspaceChooserDialog::WorkspaceChooserDialog(QWidget *parent) : QDialog(parent)
{
    setWindowTitle(tr("Select a workspace location"));
    QLabel *wpLabel = new QLabel(tr("Workspace:"), this);
    QLineEdit *wpLineEdit = new QLineEdit(QDir().homePath(), this);
    QPushButton *okButton = new QPushButton(tr("OK"), this);
    QPushButton *cancelButton = new QPushButton(tr("Cancel"), this);
    QGridLayout *layout = new QGridLayout;

    connect(okButton, SIGNAL(clicked()), this, SLOT(accept()));
    connect(cancelButton, SIGNAL(clicked()), this, SLOT(reject()));

    qDebug() << "begin: " << layout << " :end";
    layout->addWidget(wpLabel, 0, 0);
    layout->addWidget(wpLineEdit, 0, 1, 1, 2);
    layout->addWidget(okButton, 1, 1);
    layout->addWidget(cancelButton, 1, 2);
    setLayout(layout);
}

EDIT2

For reasons unknown to me, after I compiled with verbose mode -v flag set, the error didn't appear anymore, even after unsetting it again. Now gdb creates the variable, and I'm able to inspect its value.

For who are interested, the compiler's flags set are:

g++ -O0 -g3 -Wall -c -fmessage-length=0

Answer 1

Use volatile . Maybe, it'll help you!

Why do we use volatile keyword in C++?

Answer 2

What other compiler flags have you set?

Try:

-g -O0 -fno-inline

Answer 3

An easy way to force a variable to be allocated is to copy a pointer to it to a global variable.

QGridLayout **dummy; // at file scope

//...
dummy = &layout; // in function

This will force the compiler to spill the variable to memory before calling any non-inlined function, as it cannot prove that the function would not access the variable directly.

This may, of course, have performance impacts, but they're likely to be small compared to the cost of calling the GUI member functions on QGridLayout.