You're annoying for being wrong.Peter Remmers said:Am 08.04.2011 19:09, schrieb Paul:You are so annoying. You just go on and emit your attitude crap whenConsider this:
int aa[2][2]; int* p=aa[0];
...
With the above p points to an array , you think it only points to a
single int. You do not have any consideration for what is pointed to
,
you incorrectly think the pointer type defines what is pointed to.
Is it correct that in your opinion, in the above example,
the pointer p points to all of
1) aa[0][0] ,
2) aa[0] and
3) aa
?
I don't care what it points to
challenged.
That's totally irrelevant. The question's purpose was to clarify your
position. Even though you constantly promulgate your ultimate truths and
fulminate at everyone who disputes them, you can't stand by your position
when directly asked? You make me sick.
And what question can't I answer?You are unable to answer a simple question.
You evade by deflecting the question.
No I answered it correctly. You obviously have a problem with all thingsI think it poiints to a section of memory.I think it points to the int lvalue aa[0][0].
What do you think it points to?
You are unable to answer a simple question.
You evade by using dummy answers.
Sorry I dont understand your idiot mind either. You don't understand how toLeigh Johnston said:xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I do know what it points to. Saying that I am confused is a (obviously
false) premise. Kindly explain what is wrong in my logic.
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You don't know what this points to:
int* p = new int[6];
With the above p points to an array , you think it only points to a
single
int. You do not have any consideration for what is pointed to , you
incorrectly think the pointer type defines what is pointed to.
That's not the explanation, that's an assertion, and it's false.
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You have made your opinion, that a pointer-type always defines what is
pointed to, very clear.
I haven't asserted anything , you have provided this evidence. If it's
false
then it's your opinion that's false, not mine
Yes you have. All you have are assertions. Not even opinions, plain
assertions. The "Martians exist, my dad saw them!" kind.
Explain how it is possible that both pointer points to an array, and
incrementing it makes it point to the next ___element in the array__.
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Explain how what is possible?
I'm sorry but what you say is unclear.
He is quite clear. Explain how incrementing a pointer to an array will
make the pointer point to the next element in the array? It can't as it is
not a pointer to an array but a pointer to an array element; if it was a
pointer to an array incrementing it would make it point to the next array
not the next array element.
Wrong about what? About you being annoying? I don't think so.You're annoying for being wrong.Peter Remmers said:Am 08.04.2011 19:09, schrieb Paul:You are so annoying. You just go on and emit your attitude crap whenConsider this:
int aa[2][2]; int* p=aa[0];
...
With the above p points to an array , you think it only points to a
single int. You do not have any consideration for what is pointed to
,
you incorrectly think the pointer type defines what is pointed to.
Is it correct that in your opinion, in the above example,
the pointer p points to all of
1) aa[0][0] ,
2) aa[0] and
3) aa
?
I don't care what it points to
challenged.
Then why can't you answer a simple question? It was even a yes-or-noMy position cannot be any clearer, you make me sick.
You are unable to answer a simple question.And what question can't I answer?
No I answered it correctly. You obviously have a problem with all thingsI think it points to the int lvalue aa[0][0].
What do you think it points to?
I think it poiints to a section of memory.
You are unable to answer a simple question.
You evade by using dummy answers.
that are correct.
You are just wrong about everything. You are just a wrong person in allPeter Remmers said:Wrong about what? About you being annoying? I don't think so.You're annoying for being wrong.Consider this:
int aa[2][2]; int* p=aa[0];
...
With the above p points to an array , you think it only points to
a
single int. You do not have any consideration for what is pointed
to
,
you incorrectly think the pointer type defines what is pointed
to.
Is it correct that in your opinion, in the above example,
the pointer p points to all of
1) aa[0][0] ,
2) aa[0] and
3) aa
?
I don't care what it points to
You are so annoying. You just go on and emit your attitude crap when
challenged.
I did answer the question, please stop being wrong.Then why can't you answer a simple question? It was even a yes-or-no
question. Even *you* should be able to type two-letter and three-letter
words.
No you are the one who fits the adjective "dumb" here, not I.You are unable to answer a simple question.
You evade by pretending you are unaware of any question.
You are playing dumb just to be annoying, aren't you?
The only dummy here is you, the answer I gave is 100% correct.No I answered it correctly. You obviously have a problem with all thingsI think it points to the int lvalue aa[0][0].
What do you think it points to?
I think it poiints to a section of memory.
You are unable to answer a simple question.
You evade by using dummy answers.
that are correct.
You deny that it is a dummy answer?
Correct I do not understand your basic idiot mind.Leigh Johnston said:Sorry I dont understand your idiot mind either. You don't understand howxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I do know what it points to. Saying that I am confused is a
(obviously
false) premise. Kindly explain what is wrong in my logic.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You don't know what this points to:
int* p = new int[6];
With the above p points to an array , you think it only points to a
single
int. You do not have any consideration for what is pointed to , you
incorrectly think the pointer type defines what is pointed to.
That's not the explanation, that's an assertion, and it's false.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You have made your opinion, that a pointer-type always defines what is
pointed to, very clear.
I haven't asserted anything , you have provided this evidence. If it's
false
then it's your opinion that's false, not mine
Yes you have. All you have are assertions. Not even opinions, plain
assertions. The "Martians exist, my dad saw them!" kind.
Explain how it is possible that both pointer points to an array, and
incrementing it makes it point to the next ___element in the array__.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Explain how what is possible?
I'm sorry but what you say is unclear.
He is quite clear. Explain how incrementing a pointer to an array will
make the pointer point to the next element in the array? It can't as
it is not a pointer to an array but a pointer to an array element; if
it was a pointer to an array incrementing it would make it point to
the next array not the next array element.
to increment a pointer to an array?
There is not much we can do if you cannot (or will not) understand the
basics.
Hm yeah. The universal insult. That's.... childish. Try to be less childish.You are just wrong about everything. You are just a wrong person in allPeter Remmers said:Wrong about what? About you being annoying? I don't think so.Consider this:
int aa[2][2]; int* p=aa[0];
...
With the above p points to an array , you think it only points to
a
single int. You do not have any consideration for what is pointed
to
,
you incorrectly think the pointer type defines what is pointed
to.
Is it correct that in your opinion, in the above example,
the pointer p points to all of
1) aa[0][0] ,
2) aa[0] and
3) aa
?
I don't care what it points to
You are so annoying. You just go on and emit your attitude crap when
challenged.
You're annoying for being wrong.
honesty.
I did answer the question, please stop being wrong.
Hm yeah. Deflecting adjectives. That's... childish. Try to be less childish.No you are the one who fits the adjective "dumb" here, not I.
The only dummy here is you, the answer I gave is 100% correct.I think it points to the int lvalue aa[0][0].
What do you think it points to?
I think it poiints to a section of memory.
You are unable to answer a simple question.
You evade by using dummy answers.
No I answered it correctly. You obviously have a problem with all things
that are correct.
You deny that it is a dummy answer?
Explain how it is possible that both pointer points to an array, and
incrementing it makes it point to the next ___element in the array__.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Explain how what is possible?
I'm sorry but what you say is unclear.
Explain how it is possible that both pointer points to an array, and
incrementing it makes it point to the next ___element in the array__.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Explain how what is possible?
I'm sorry but what you say is unclear.
No, it's clear enough, and you are lying when saying it's not. If you
say that pointer of type "T*" (somehow, sometime, whatever) points to
an __array__ of T, then simplest of pointer operations (indexing,
incrementing) are broken.
Since pointer operations clearly aren't broken, your claim is false.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
What is broken about indexing and incrementing?
No, it's clear enough, and you are lying when saying it's not. If you
say that pointer of type "T*" (somehow, sometime, whatever) points to
an __array__ of T, then simplest of pointer operations (indexing,
incrementing) are broken.
Since pointer operations clearly aren't broken, your claim is false.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
What is broken about indexing and incrementing?
You create a 2d array, then you create a pointer to a 1d-subarray of this2d
array.
Then after that I don't know what you are trying to do.
No, it's clear enough, and you are lying when saying it's not. If you
say that pointer of type "T*" (somehow, sometime, whatever) points to
an __array__ of T, then simplest of pointer operations (indexing,
incrementing) are broken.Since pointer operations clearly aren't broken, your claim is false.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
What is broken about indexing and incrementing?
int a[2][2]={1,2,3,4};
int* p1 = a[0]; // p1 points to an array (your claim).
p1++;
if (*p1 != a[1])
std::cout << "incrementing is broken.";
int* p2 = a[0]; // p2 points to an array (your claim).
if (*p2[1] != a[1])
std::cout << "indexing is broken.";
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You create a 2d array, then you create a pointer to a 1d-subarray of this2d
array.
Then after that I don't know what you are trying to do. You say its broken
but you don't say what you expect it to do.
No, it's clear enough, and you are lying when saying it's not. If you
say that pointer of type "T*" (somehow, sometime, whatever) points to
an __array__ of T, then simplest of pointer operations (indexing,
incrementing) are broken.Since pointer operations clearly aren't broken, your claim is false.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
What is broken about indexing and incrementing?
int a[2][2]={1,2,3,4};
int* p1 = a[0]; // p1 points to an array (your claim).
p1++;
if (*p1 != a[1])
std::cout << "incrementing is broken.";
int* p2 = a[0]; // p2 points to an array (your claim).
if (*p2[1] != a[1])
std::cout << "indexing is broken.";
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You create a 2d array, then you create a pointer to a 1d-subarray of this
2d
array.
Then after that I don't know what you are trying to do. You say its broken
but you don't say what you expect it to do.
you're so confused it is comical.
"Goran" <[email protected]> wrote in messageint a[2][2]={1,2,3,4};
int* p1 = a[0]; // p1 points to an array (your claim).
p1++;
if (*p1 != a[1])
std::cout << "incrementing is broken.";
int* p2 = a[0]; // p2 points to an array (your claim).
if (*p2[1] != a[1])
std::cout << "indexing is broken.";xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You create a 2d array, then you create a pointer to a 1d-subarray of this
2d
array.
Then after that I don't know what you are trying to do. You say its broken
but you don't say what you expect it to do.
I expect both pointers to point to a[1]. This is because you say that
they point to one array (a[0]), and I know that incrementing, or
indexing pointers with 1, must make them point to next element. Thats
a[1]. This is how pointers work. Explain why this doesn't happen.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Incrementing it will make it point to a[0][1] becuase you increment a
pointer to a 1d array.
[...]
If p is a pointer to a 1d array, how come I can assign address
of a single int to it? a[0][1] and i are clearly not arrays.
int a[2][2]={1,2,3,4};
int* p1 = a[0]; // p1 points to an array (your claim).
p1++;
if (*p1 != a[1])
std::cout << "incrementing is broken.";
int* p2 = a[0]; // p2 points to an array (your claim).
if (*p2[1] != a[1])
std::cout << "indexing is broken.";xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You create a 2d array, then you create a pointer to a 1d-subarray of
this
2d
array.
Then after that I don't know what you are trying to do. You say its
broken
but you don't say what you expect it to do.
I expect both pointers to point to a[1]. This is because you say that
they point to one array (a[0]), and I know that incrementing, or
indexing pointers with 1, must make them point to next element. Thats
a[1]. This is how pointers work. Explain why this doesn't happen.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Incrementing it will make it point to a[0][1] becuase you increment a
pointer to a 1d array.
No I mean p points, primarily to an array.SG said:[...]
If p is a pointer to a 1d array, how come I can assign address
of a single int to it? a[0][1] and i are clearly not arrays.
When he says "p is a pointer to a 1d array" he means "p currently
stores an address of an array or the address of something that is part
of that array".
A pointer stores an address of an entity. A pointer needs to be treated asSo, when he says "p is ..." he's mainly considering
runtime properties and the raw type-agnostic address some object
pointer stores.
No I acknowledge what type of pointers I use.Unlike others, he ignores the type of a pointer
variable when it comes to describing what a certain pointer "is".
That's why he brought up the broken finger/bunch/banana analogy.
Like you obviously do?Guys, don't you realize that this is going around in circles? Don't
you have anything better to do?
No, it's clear enough, and you are lying when saying it's not. If you
say that pointer of type "T*" (somehow, sometime, whatever) points to
an __array__ of T, then simplest of pointer operations (indexing,
incrementing) are broken.
Since pointer operations clearly aren't broken, your claim is false..
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
What is broken about indexing and incrementing?
int a[2][2]={1,2,3,4};
int* p1 = a[0]; // p1 points to an array (your claim).
p1++;
if (*p1 != a[1])
std::cout << "incrementing is broken.";
int* p2 = a[0]; // p2 points to an array (your claim).
if (*p2[1] != a[1])
std::cout << "indexing is broken.";
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You create a 2d array, then you create a pointer to a 1d-subarray of
this
2d
array.
Then after that I don't know what you are trying to do. You say its
broken
but you don't say what you expect it to do.I expect both pointers to point to a[1]. This is because you say that
they point to one array (a[0]), and I know that incrementing, or
indexing pointers with 1, must make them point to next element. Thats
a[1]. This is how pointers work. Explain why this doesn't happen.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Incrementing it will make it point to a[0][1] becuase you increment a
pointer to a 1d array.
OK. So, I'll presume then this claim is true, that int* is a pointer
to 1d array. If so (continuing from previous example)...
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Its not a claim its a fact.
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No, it's clear enough, and you are lying when saying it's not. If
you
say that pointer of type "T*" (somehow, sometime, whatever) points
to
an __array__ of T, then simplest of pointer operations (indexing,
incrementing) are broken.
Since pointer operations clearly aren't broken, your claim is false.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
What is broken about indexing and incrementing?
int a[2][2]={1,2,3,4};
int* p1 = a[0]; // p1 points to an array (your claim).
p1++;
if (*p1 != a[1])
std::cout << "incrementing is broken.";
int* p2 = a[0]; // p2 points to an array (your claim).
if (*p2[1] != a[1])
std::cout << "indexing is broken.";
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
You create a 2d array, then you create a pointer to a 1d-subarray of
this
2d
array.
Then after that I don't know what you are trying to do. You say its
broken
but you don't say what you expect it to do.I expect both pointers to point to a[1]. This is because you say that
they point to one array (a[0]), and I know that incrementing, or
indexing pointers with 1, must make them point to next element. Thats
a[1]. This is how pointers work. Explain why this doesn't happen.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Incrementing it will make it point to a[0][1] becuase you increment a
pointer to a 1d array.
OK. So, I'll presume then this claim is true, that int* is a pointer
to 1d array. If so (continuing from previous example)...
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Its not a claim its a fact.
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