character pointer worries

this may sound strange, but each time i try to print the address of a character, it prints out the character.

for example,

Code:

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    char c='a',*p;
    p=&c;
    cout<<p;

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gives the output 'a'.

i can get over it using void pointers, but why does the above happen and is there any other way out?

I think the second line should be:

Code:

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*p = &c;

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Try that.

Quote:

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Originally posted by kiss_psycho
why does the above happen and is there any other way out?

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That's usually what you want to happen - after all, we use char *'s for C-style strings, so that's how cout handles it. Casting to a void pointer should be just fine.

oh no josh, &c gives an address which can only be stored in a character type pointer. the syntax i wrote is absolutely correct, there must be some problem with the character representation in C++ since i have used tc++ v3.0, borland c++, and vc++ 6.0, all to no avail - they all give the same result.

but omnius, & should produce an address. strange, isnt there any explainable reason behind this?

Quote:

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Originally posted by kiss_psycho
but omnius, & should produce an address. strange, isnt there any explainable reason behind this?

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It does produce an address ... of a character. That's what variable p contains. In C++ the output stream handles this as a C-string. That's all there is to it. If you're still not convinced try some simple exercises using C-style strings and pointers and you'll realise what's going on.

:eek: thanks omni, guess i got confused.

Quote:

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Originally posted by joshdick
I think the second line should be:

Code:

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*p = &c;

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Try that.

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That's incorrect. The problem is, when you pass the address of the character to cout, it thinks you are handing it an array of chars, and so tries to print a string. The same effect if the OP had done:

cout<<&c;

To get around that, you could do:

cout<<(int)p;

That way, cout interprets the address as an integer. To view the out put in hex I think you can do:

cout<<ios::hex<<(int)p;

Quote:

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Originally posted by kiss_psycho
:eek: thanks omni, guess i got confused.

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That's understandable. Remember that the << operator is overloaded for all of the fundamental types such as int, double, and char *. You were passing it a char *.