Re: How printf() works???????

On 7 Mar, 09:10, "Robbie Hatley" <lonew...@xxxxxxxx> wrote:
"Richard" <de...@xxxxxxxxx> wrote:
Richard Heathfield <r...@xxxxxxxxxxxxxxx> writes:

sant.ta...@xxxxxxxxx said:

Hello,

I would appreciate some comments about the piece of code given below
and explanation about the result of this specified code.

int x = 20;

printf("%d %d %d",x < 30, x = 40; x > 10);

The output of these code will be :: 0 40 1

Well, it might be, once you wrap a program around it and fix the syntax
error (a semicolon instead of a comma). First, I'll explain why you might
get that output, and then I'll explain why you might not.

x < 30 is a relational expression, and all expressions have values.
Relational expressions have the value 0 if they're false, and 1 if they're
true. So x < 30 will evaluate to 0 if x is less than 30, and 1
otherwise.

It won't actually.

How do you figure?  It does in standard C.  If it doesn't for you,
then your compiler is broken and you should get a better one.


Richard Heathfield made a typo. He meant "So x < 30 will evaluate to
*1* if x
is less than 30, and *0* otherwise."



--
Nick Keighley



.

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