order of execution

In what order will the following line of code be executed?

Code:

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while((char c = cin.get()) != 'q')

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Is my assumption correct that the sequence of execution will be in the following order.

1) cin.get() gets called
2) its return value is assigned to c
3) then c's value is compared to 'q'

Is it correct? I was reading Thinking in C++ by Bruce Eckel and in the third Chapter I read this

Excerpt from the book:

Quote:

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For example, you cannot have any parentheses. That is, you cannot say:

while((char c = cin.get()) != 'q')

The addition of the extra parentheses would seem like an innocent and useful thing to do, and because you cannot use them, the results are not what you might like. The problem occurs because ‘!=’ has a higher precedence than ‘=’, so the char c ends up containing a bool converted to char. When that’s printed, on many terminals you’ll see a smiley-face character.

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You get a syntax error is what you get. If you put the declaration of c outside the while statement, then you'd be fine -- but you can't nest a declaration of a variable like that.

Quote:

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Originally Posted by tabstop

You get a syntax error is what you get. If you put the declaration of c outside the while statement, then you'd be fine -- but you can't nest a declaration of a variable like that.

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A what? Clarify please

A syntax error. What you have typed is not legal, valid, C++. You can do this:

Code:

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while((c = cin.get()) != 'q') //fine

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no problem. But you cannot nest a variable declaration inside extra parentheses as you have tried to do above.

How come this statement is valid?

Code:

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while(char c = cin.get() != 'q')

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What complier does your book tell you to use? In Microsoft Visual C++ (MSVC) you can not declare a variable inside the while's condition like that.

So you mean that (my last-most post's code) is not valid?

It is not.

Quote:

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Originally Posted by manzoor

How come this statement is valid?

Code:

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while(char c = cin.get() != 'q')

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It is syntactically correct, yes. It is logically incorrect, as the != will happen first and so char c will be assigned either one or zero, not the result of cin.get(), but it will compile (unlike the first example).

Code:

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while(char c = cin.get() != 'q')

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You can hot have declare a variable in there. You are trying to declare c in there with char.

Code:

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char c;
while((c = cin.get()) != 'q')

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You have to declare the variable outside.

Atlast, the day is saved by PowerPuff Girls, la la laaaaaaaaaaaaa


Thanks guys, this forum has always been a helpful and a good learning resource

Quote:

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Originally Posted by CrissyCrisCris

Code:

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while(char c = cin.get() != 'q')

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You can hot have declare a variable in there. You are trying to declare c in there with char.

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Yes you can. That's syntactically correct code. You could have at least tried it before expounding false things.

Quote:

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Originally Posted by brewbuck

Yes you can. That's syntactically correct code. You could have at least tried it before expounding false things.

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I believe CrissyCrisCris did?

Quote:

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Originally Posted by CrissyCrisCris

What complier does your book tell you to use? In Microsoft Visual C++ (MSVC) you can not declare a variable inside the while's condition like that.

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Quote:

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Originally Posted by Elysia

I believe CrissyCrisCris did?

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This compiles with MSVC8:

Code:

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char foo()
{
    return 0;
}

int main()
{
    while(char c = foo() != 'q');
}

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Ah, mistake on his/her part then, or perhaps something else.
Syntactically correct, semantically incorrect. So nice.

What the book means is that you cannot have both the char in there and the extra brackets (which are needed to make it make sense). You have sort of cut out the context:

Quote:

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Although the example* also shows variables defined within while, if, and switch statements, this kind of definition is much less common than those in for expressions, possibly because the syntax is so constrained. For example, you cannot have any parentheses.

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* This refers to a previous example.