m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
What does this mean?
I have seen v=&var->member.thing ;
but what does it mean when you change the & for int?
-
m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
What does this mean?
I have seen v=&var->member.thing ;
but what does it mean when you change the & for int?
-
Re: What does this mean?
JoeC wrote:You don't change anythign for anything. The unary operator & ism_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
>
What does this mean?
>
I have seen v=&var->member.thing ;
>
but what does it mean when you change the & for int?
for taking the address of the operand. The (blah) notation is
for *casting* the left operand into a different type. Look it up
in your favourite C++ book (BTW, which one are you reading that
doesn't describe those?)
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
-
Re: What does this mean?
JoeC wrote:>
m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
>
What does this mean?
>
I have seen v=&var->member.thing ;
>
but what does it mean when you change the & for int?
v=&var->member.thing ; means that v is going to get the address of
var->member.thing .
&'s precedence is lower than both -and .
Think of it as:
v = &(var->member.thing )
v must be a pointer to whatever type var->member.thing is.
The change you are asking about wouldn't be all, because the type of v
would have to change too.
If this compiles:
int *v = &var->member.thing ;
this of course won't:
int *v = (int)var->member.thing ;
but this will:
int v = (int)var->member.thing ;
LR
Comment
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Re: What does this mean?
On Sep 18, 3:24 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:The (int) casts what's to the right of it (I'm sureJoeC wrote:>m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
The unary operator & is
for taking the address of the operand. The (blah) notation is
for *casting* the left operand into a different type.
you know this, but you wrote 'left operand').
Also, in this example it casts the entirety of the
right-hand side of the '=', since -and . bind
more tightly than the cast.
Comment
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Re: What does this mean?
"JoeC" <enki034@yahoo. comwrote in message
news:1190080632 .256341.38700@n 39g2000hsh.goog legroups.com...To restate what Victor said,>
m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
>
What does this mean?
>
I have seen v=&var->member.thing ;
>
but what does it mean when you change the & for int?
(int) is a "c style cast". In C++ we could use that or the safer, prefered
static_cast<int >.
& is "address of"
* is "contents of" (also called dereferencing).
A cast will convert one type to another. A static cast generally is used to
covert the value of one type to another. Say, for instance, biWidth was a
double. The (int) says to convert it to an integer.
Comment
-
Re: What does this mean?
Old Wolf wrote:It was a braino, I was thinking that both operators are writtenOn Sep 18, 3:24 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:>>JoeC wrote:>>>> m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
>> m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
>The unary operator & is
>for taking the address of the operand. The (blah) notation is
>for *casting* the left operand into a different type.
The (int) casts what's to the right of it (I'm sure
you know this, but you wrote 'left operand').
to the left of what they operate on... Thanks for noticing and
correcting.
I am guesing you use the terms "bind more tightly" in place ofAlso, in this example it casts the entirety of the
right-hand side of the '=', since -and . bind
more tightly than the cast.
"have higher precedence".
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Comment
-
Re: What does this mean?
Jim Langston wrote in message...prefered"JoeC" wrote in message...>
m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;
What does this mean?
I have seen v=&var->member.thing ;
but what does it mean when you change the & for int?
To restate what Victor said,
>
(int) is a "c style cast". In C++ we could use that or the safer,tostatic_cast<int >.
& is "address of"
* is "contents of" (also called dereferencing).
>
A cast will convert one type to another. A static cast generally is usedAlso:covert the value of one type to another. Say, for instance, biWidth was a
double. The (int) says to convert it to an integer.
>
// C style cast
m_iWidth = (int) pBitmapInfo->bmiHeader.biWi dth;
// C++ style (conversion. watch your compile 'warnings'.)
m_iWidth = int( pBitmapInfo->bmiHeader.biWi dth );
--
Bob R
POVrookie
Comment
-
Re: What does this mean?
On Sep 17, 10:24 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:What I mean is that where the (int) is there is a&. That is I can getJoeC wrote:>m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;>What does this mean?>I have seen v=&var->member.thing ;>but what does it mean when you change the & for int?
You don't change anythign for anything. The unary operator & is
for taking the address of the operand. The (blah) notation is
for *casting* the left operand into a different type. Look it up
in your favourite C++ book (BTW, which one are you reading that
doesn't describe those?)
>
data from a pointer but what does the (int) or the type in () mean?
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Comment
-
Re: What does this mean?
On Sep 17, 10:55 pm, LR <lr...@superlin k.netwrote:Thanks, it makes sense now.JoeC wrote:
>>m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;>What does this mean?>I have seen v=&var->member.thing ;>but what does it mean when you change the & for int?
v=&var->member.thing ; means that v is going to get the address of
var->member.thing .
>
&'s precedence is lower than both -and .
Think of it as:
v = &(var->member.thing )
>
v must be a pointer to whatever type var->member.thing is.
>
The change you are asking about wouldn't be all, because the type of v
would have to change too.
>
If this compiles:
int *v = &var->member.thing ;
this of course won't:
int *v = (int)var->member.thing ;
but this will:
int v = (int)var->member.thing ;
>
>
LR
Comment
-
Re: What does this mean?
JoeC wrote in message...It's a 'cast'. Remove it and re-compile. Does the compiler complain?On Sep 17, 10:24 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:What I mean is that where the (int) is there is a&. That is I can getJoeC wrote:m_iWidth = (int)pBitmapInf o->bmiHeader.biWi dth;
m_iHeight = (int)pBitmapInf o->bmiHeader.biHe ight;What does this mean?I have seen v=&var->member.thing ;You don't change anythign for anything. The unary operator & isbut what does it mean when you change the & for int?
for taking the address of the operand. The (blah) notation is
for *casting* the left operand into a different type. Look it up
in your favourite C++ book (BTW, which one are you reading that
doesn't describe those?)
data from a pointer but what does the (int) or the type in () mean?
>
--
Bob R
POVrookie
Comment
Comment