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what do you think "const" mean ? constipated ?
for me (and for C/C++ also), it means "constant", that is, you cannot modify it ! that's all....
TOXCCT >>> GEII power
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Yes of course I know this (I'm not completely stupid  ) but if you try to change the contents of it, you will have a compile error.
Or here, you change it by another way (using the adress of the variable). And there is no errors BUT the adress p and the adress &a are the same but their contents are different so there is something strange (two pointers that are pointing at exactly the same adress but the values they hold are different  )!!
It would be normal if *p still contain 10 at the end (that menas the content of a was unchanged)
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in fact, the compiler don't say anything because const is an identifier for the compiler, and as we don't explicitely write into variable a, there is no compile error. but a is still const (certainly not even in the stack...) so *p cannot access the value correctly to alter it.
TOXCCT >>> GEII power
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Sounds like your constipated!!!!!
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hey man, first, you write as an anonymous, and secondly, where did you sense of humour go ???
did you trash it ? 
TOXCCT >>> GEII power
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Sounds like your are constipated!!!!
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pouet pouet pouet traallalalilouuu
TOXCCT >>> GEII power
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if you do this:
int a = 10;
=> a is the value AND &a is the pointer
int* p = 100;
=> *p is the value AND p is the pointer
thus if I see correctly: int* p=(int*)&a; means that *p = 10 (* and & are opposites) and p and &a are still different. (you should do p = &a; if you would like to copy pointers)
Then again I could be mistaken (you always do working with pointers)
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excuse me but int* p=(int*)&a; is as much correct as your p = &a;
secondly, you say
int a = 10;
a is the value AND &a is the pointer. a say NO. &a is the ADDRESS !!
and to finish, even if you're right, you don't answer the post to typed...
TOXCCT >>> GEII power
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If you try to do
p = &a;
in this example the compiler will complain that it can not convert from const int* to int* ! 
Ant.
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V. wrote:
int* p = 100;
=> *p is the value AND p is the pointer
Yes but he is assigning the adress of a in p just before:
int* p=(int*)&a;<br />
So p and &a are the same (in this case, the adress of &a).
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and if you output *(int*)&a rather than a? 
Ant.
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where do you see a difference between *(int*)&a and a you, except in confusing the reader ?? 
TOXCCT >>> GEII power
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Try it in VC 6 you will be amazed, anyway see my other post WRT casting away const-ness.
Ant.
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To all:
const int a=10;//Declare an int constant and init 10
int* p=(int*)&a; //Declare an int pointer p and point to address of a.
*p=100;//Change where p pointed to 100
int i=a;//Declare an int variable i and init i to value of a.
cout << "p=" << p << endl;//Print p value (address)
cout << "&a=" << &a << endl;//Print address of a
cout << "*p=" << *p << endl;//Print value at where p pointed to
cout << "a=" << a << endl;//Print value of a
cout << "i=" << i << endl;//Print value of i
And the result:
p=0012FF7C //Point to the same address of a
&a=0012FF7C //Point to the same address of p
*p=100 //But one value is 100
a=10 //And here the other value is 10
i=10 //And here the value is 10 too.
p and a have the same memory address. But it's value is difference. Why that?
PS: if I write: int* p=&a, compiler will compain as an error. So I have used a trick by casting int* p=(int*)&a. In this case, compiler ignore.
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Well it gives a==100 on my compiler! But i==10. I'd imagine that the problem was set to explain why i != a.
You're right that the compiler effectively replaces all references to a with the value 10, and that's why i == 10 (because the compiler sees this line as being i = 10) but the value a does have a physical address and using the two lines:
<br />
int* p=(int*)&a;<br />
*p=100;<br />
should change a.
If I was you I'd debug the code and then look at the disassembly, there everything should become clear. In my compiler it gives this:
<br />
1844: const int a=10;<br />
0050D1BF mov dword ptr [a],0Ah<br />
1845: int* p=(int*)&a;<br />
0050D1C6 lea eax,[a]<br />
0050D1C9 mov dword ptr [p],eax<br />
1846: *p=100;<br />
0050D1CC mov ecx,dword ptr [p]<br />
0050D1CF mov dword ptr [ecx],64h<br />
1847: int i=a;<br />
0050D1D5 mov dword ptr [i],0Ah<br />
Here it's clear that a will be changed to 100 in line 1846 and also that i will be set to 10 in line 1847.
By the way, const isn't magic, it just tells the compiler that you shouldn't be able to change a using a line like a = 20; it doesn't add some kind of 'lock' to a meaning it can never change.
Oh yeah, I've just spotted why a is 10 in your output. Because you output a which has been replaced at compile time with 10!!
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Finally it's clear  !!
Yes ok, I didn't know that the compiler really replace all occurences of a by it's value (so acting like a #define in fact).
What did you win ??
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Yea it seems that the compiler has optimised itself here since a was declared const! 
Ant.
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good catch! here: ![Good Stuff | [The good stuff]](https://codeproject.global.ssl.fastly.net/script/Forums/Images/vegemite.gif) enjoy on me
~Nitron.
ññòòïðïðB A
start
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nguyenvhn wrote:
const int a=10;
This simply means that you cannot modify memory address 0x0012FF7C through variable a. The address is not read-only, the variable is. make sense?
"The pointy end goes in the other man." - Antonio Banderas (Zorro, 1998)
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teu teu teu, read the other posts before saying so, especially "Christopher Lloyd"'s one
TOXCCT >>> GEII power
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In which case he states that "it just tells the compiler that you shouldn't be able to change a using a line like a = 20" In my post, I stated that "you cannot modify memory address 0x0012FF7C through variable a"
What's your point?
"The pointy end goes in the other man." - Antonio Banderas (Zorro, 1998)
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yes you can
const is only a specification for the compiler!!!
if you had read all the posts here, you would have seen that we already said that also... and i even that i was thinking like you ! 
TOXCCT >>> GEII power
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Then how would you propose changing the value through variable a?
"The pointy end goes in the other man." - Antonio Banderas (Zorro, 1998)
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