Will the second be evaluated if the first is NULL?

[William Payne]

Consider the following code:

struct foo
{
char* bar;
};

// fooptr is of type foo*
if(fooptr != NULL && fooptr->bar != NULL)
{
cout << fooptr->bar << endl;
}

Is the if-statement safe, meaning it won't try to evaluate fooptr->bar if
fooptr is NULL? Can I make sure it is evaluated from left to right or should
I have nested if-statements instead?

Thanks for any replies

/ William Payne

 

[Karl Heinz Buchegger]

Consider the following code:

struct foo
{
char* bar;
};

// fooptr is of type foo*
if(fooptr != NULL && fooptr->bar != NULL)
{
cout << fooptr->bar << endl;
}

Is the if-statement safe, meaning it won't try to evaluate fooptr->bar if
fooptr is NULL? Can I make sure it is evaluated from left to right or should
I have nested if-statements instead?

In x && y, it is guaranteed that y is not evaluated if x is already false.
Same for x || y : if x is already true, y is not evaluated at all.

Which books are you using, that do not mention 'shortcut evaluation' ?

 

[Julie]

Consider the following code:

struct foo
{
char* bar;
};

// fooptr is of type foo*
if(fooptr != NULL && fooptr->bar != NULL)
{
cout << fooptr->bar << endl;
}

Is the if-statement safe, meaning it won't try to evaluate fooptr->bar if
fooptr is NULL? Can I make sure it is evaluated from left to right or should
I have nested if-statements instead?

You're golden -- yes, the statement is safe.

The behavior is called 'short circuit', and evaluated left to right.

 

[William Payne]

In x && y, it is guaranteed that y is not evaluated if x is already false.
Same for x || y : if x is already true, y is not evaluated at all.

Which books are you using, that do not mention 'shortcut evaluation' ?

Thanks for the quick reply, Karl. No need for a nested if-statement then,
just as I thought. And to answer your question: I don't have access to a C++
text except Josuttis Standard Library Book at the moment. =/

/ William Payne