pointers of chars and arrays of chars Question

hi

i noticed something while working with arrays, if i do this:

Code:

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int main()
{
    char* a="abc";

    printf("%p %p %p %p", a, &a[0], &a[1], &a[2]);

    a="cba";

    printf("\nsecond values: %p %p %p %p", a, &a[0], &a[1], &a[2]);
}

---

the values in the "second values" are diferent, meaning that a[0] changed its position.

so this means that everytime i change an array like this: a="HELLO WORLD" the a[0] will change its position??

a isn't an array. It's a pointer. You are indexing into whatever memory location it points to. So when you change where it points to, then the offsets from that base will also likely be different (IE: &a[0], &[a1], etc..).

So that means that if i have a="b" that would be like making 'a' point to the address number "b" (witch i think it gives a=4206666 because i casted it to an int)??

Quote:

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Originally Posted by shiroaisu

So that means that if i have a="b" that would be like making 'a' point to the address number "b" (witch i think it gives a=4206666 because i casted it to an int)??

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You aren't casting a char pointer to an int. A pointer is a 32 bit (can be 64 on x64 systems??? donno for sure) integer to begin with. Just because it is a CHAR pointer, doesn't mean the variable/object is still a CHAR. It is still a pointer. It just gets interpreted as that object type if you were to do something like cout<< (*a) <<endl; and it also tells how many bytes the memory location is.

Try this once...

Code:

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cout<< sizeof(char) <<endl;
cout<< sizeof(char*) <<endl;
cout<< sizeof(int) <<endl;
cout<< sizeof(int*) <<endl;
cout<< sizeof(double) <<endl;
cout<< sizeof(double*) <<endl;

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Pointers are memory locations. That 0x04206666 value you saw is a location in memory.

Quote:

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Originally Posted by scwizzo

You aren't casting a char pointer to an int. A pointer is a 32 bit (can be 64 on x64 systems??? donno for sure) integer to begin with. Just because it is a CHAR pointer, doesn't mean the variable/object is still a CHAR. It is still a pointer. It just gets interpreted as that object type if you were to do something like cout<< (*a) <<endl; and it also tells how many bytes the memory location is.

Try this once...

Code:

---

cout<< sizeof(char) <<endl;
cout<< sizeof(char*) <<endl;
cout<< sizeof(int) <<endl;
cout<< sizeof(int*) <<endl;
cout<< sizeof(double) <<endl;
cout<< sizeof(double*) <<endl;

---

Pointers are memory locations. That 0x04206666 value you saw is a location in memory.

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yes, i know all of that, what i meant to say is that i did like this: int k="b" and it gave me that number 4206666
so doing char* a="b" would be the same that char* a=4206666 right?

Argh, I'm wrong. Your pointer should be fine... Forget my nonsense.

The reason it's different is because you are setting the pointer to point to "abc" and then to "cba" which are different string literals at different locations in the program's memory. So, naturally, they are different memory addresses. In fact, all of the positions should be different in the second case.

Quote:

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Originally Posted by scwizzo

You aren't casting a char pointer to an int. A pointer is a 32 bit (can be 64 on x64 systems??? donno for sure) integer to begin with. Just because it is a CHAR pointer, doesn't mean the variable/object is still a CHAR. It is still a pointer. It just gets interpreted as that object type if you were to do something like cout<< (*a) <<endl; and it also tells how many bytes the memory location is.

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Naturally. Otherwise it can't address >4 GB of memory.
Also, are you doing C or C++?
int k="b" will not compile with C++. Nor should you be using printf for that matter.

Quote:

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Originally Posted by syzygy

Argh, I'm wrong. Your pointer should be fine... Forget my nonsense.

The reason it's different is because you are setting the pointer to point to "abc" and then to "cba" which are different string literals at different locations in the program's memory. So, naturally, they are different memory addresses. In fact, all of the positions should be different in the second case.

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but w8, that means that the pointers location, the location of a, will change acording to the inputed string??

and btw im doing c now.

No, a string literal is stored in a far-away place and when you try to reference it, it decays to a pointer.
When you input data, you are writing data to a memory location.
String literals are already written somewhere. You just take the address where they are stored.

And if you are doing C, perhaps you shouldn't post in the C++ forum?

Quote:

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Originally Posted by Elysia

No, a string literal is stored in a far-away place and when you try to reference it, it decays to a pointer.
When you input data, you are writing data to a memory location.
String literals are already written somewhere. You just take the address where they are stored.

And if you are doing C, perhaps you shouldn't post in the C++ forum?

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ok then, i think i understand what happens now.

as for the forum, well, at first i didnt realized that i was posting in this forum so it was a mistake, but then i thought that maybe it isnt so bad because the subject is pointers and ill move on to c++ soon so ppl would also give me warnings like the ones that u gave me about int b="o" not working for example