whats char*?

i'v been reading the tutorials on this website until i got to lesson 9: C Strings. it mentions that single characters cannot be passed into functions. the function is expecting char*. what is that? i'v read all the tutorials before it and am pretty sure that none of them mention anything about char*. i know from the pointers tutorial that *char is used to declare pointers, and to access the actual memory address. so what does char* do?

Your pointer tutorial, or at least your understanding of it, is backwards. There is no '* char'. It's 'char *'. Read it right to left. "Pointer to a char".


Quzah.

I find a lot of new programmers forget it's char* not *char because they're so used to typing

Code:

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char *myVar;

/* not */

char* myVar;

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That's just what I've observed, though.

Quote:

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Originally Posted by SlyMaelstrom

I find a lot of new programmers forget it's char* not *char because they're so used to typing

Code:

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char *myVar;

/* not */

char* myVar;

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That's just what I've observed, though.

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Whitespace doesn't matter. You can even do:

Code:

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char*foo;

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The * ties to the variable, not the type, so this:

Code:

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char* foo, bar;

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only gives you 1 pointer and 1 char. It doesn't make 'foo' and 'bar' pointers. For that you'd need something like:

Code:

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char *foo, *bar;

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Again, white space doesn't matter. It could have been:

Code:

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char*foo,*bar;
char* foo, *bar;
char *foo,* bar;
char * foo, * bar;

---

Quzah.

I think you're missing my point. I know how the operator works... what I'm saying is new programmers are so used to typing it directly preceding a word rather than following a word, that when they simply write the data type char* without an identifier, they accidentally write it *char. You see?

There were several people in the lower level programming courses that I took who would write a pointer declaration fine, but when it came to prototyping, they'd often have

Code:

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void myFunc(const *char);

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...or something of that nature. Again, just my observation... I don't know if that's the real reason it's done as often as it is.

ok so for declairing a pointer i type

Code:

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int *pointer

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for example and to access the actual memory address i use

Code:

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printf("%d", pointer*);

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is that right?

Quote:

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Originally Posted by Abda92

i'v been reading the tutorials on this website until i got to lesson 9: C Strings. it mentions that single characters cannot be passed into functions. the function is expecting char*. what is that? i'v read all the tutorials before it and am pretty sure that none of them mention anything about char*. i know from the pointers tutorial that *char is used to declare pointers, and to access the actual memory address. so what does char* do?

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Hmmm, while getting twisted knickers about your typo nobody has addressed your point so far, have they? I think the wording of lesson 9 has misled you, you certainly are able to pass a single character to a function and if the function declaration specifies a char data type rather than a char * data type it will be expecting you to pass a single character. Mistakes are made when people think that "a" is a single character. "a" in C is actually a string, comprising a byte with the value of the character 'a' followed by a byte with the value zero. If you want to pass the single character you must pass either the character literal 'a' or a char data type variable.

Quote:

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Originally Posted by lesson 9

Remember how arrays act like pointers when passed into functions? Characters don't, so if you pass a single character into a function, it won't work; the function is expecting a char*, not a char.

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This should say "so if you pass a single character into a function which is expecting a string, it won't work".

Quote:

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Originally Posted by Abda92

ok so for declairing a pointer i type

Code:

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int *pointer

---

for example and to access the actual memory address i use

Code:

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printf("%d", pointer*);

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is that right?

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Abda92 == "100% Confused"

No, you would printf("%d", *pointer);

Please notice the difference between a data type and an identifier.

Quote:

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Originally Posted by Abda92

ok so for declairing a pointer i type

Code:

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int *pointer

---

for example and to access the actual memory address i use

Code:

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printf("%d", pointer*);

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is that right?

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No, it's

Code:

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int teajerr = 9;
int * p = &teajerr;

print ("%d\n", *p);

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Quote:

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Originally Posted by risby

Hmmm, while getting twisted knickers about your typo nobody has addressed your point so far, have they?

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Quote:

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Originally Posted by risby

If you want to pass the single character you must pass either the character literal 'a' or a char data type variable.

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Or the address of said variable, ie, a pointer to a character.

Code:

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void foo( char * bar );
...

char bar;
...
foo( &bar );

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Quote:

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Originally Posted by risby

This should say "so if you pass a single character into a function which is expecting a string, it won't work".

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Sure it can.

Code:

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char foo = '\0';
size_t x;
...
x = strlen( &foo );

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That'll work.

See, if you're going to try to correct people, be correct. ;)

You can pass single characters to functions which take pointers to characters, provided you've actually got the variable itself, and not just the value. If it "works" or not, depends on the value of the variable, and the function. It's usually not a good idea to pass single characters to all of the "string" functions. But it can be done.


Quzah.

Quote:

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Originally Posted by quzah

See, if you're going to try to correct people, be correct. ;)

You can pass single characters to functions which take pointers to characters,

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I don't think your comment helps Quzah. You are not passing "single characters to a function which takes pointers to character" if you pass the address of a single character; you are passing a pointer.

ok i took this piece of code from the tutorial and compiled it, and it worked! but it doesn't use p*, it uses*p

Code:

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#include <stdio.h>

int main()
{
    int x;            /* A normal integer*/
    int *p;          /* A pointer to an integer ("*p" is an integer, so p
                      must be a pointer to an integer) */

    p = &x;          /* Read it, "assign the address of x to p" */
    scanf( "%d", &x );          /* Put a value in x, we could also use p here */
    printf( "%d\n", *p ); /* Note the use of the * to get the value */
    getchar();
}

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EDIT: sorry, the page didn't refresh until i postid this, ignore it

Abda... a little bit of reading into what has been said in this post would have solved your questions hours ago.

Quote:

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Originally Posted by Abda92

i know from the pointers tutorial that *char is used to declare pointers

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Quote:

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Originally Posted by Abda92

ok i took this piece of code from the tutorial and compiled it, and it worked! but it doesn't use p*, it uses*p

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p* is not the way to dereference a pointer, *p is.

* char is not the way to declare a pointer, char * is.

so what is the difference between data types and an identifier? i'm sorry i really am confused:confused:

Quote:

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Originally Posted by risby

p* is not the way to dereference a pointer, *p is.

* char is not the way to declare a pointer, char * is.

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i'm sorry i think i'm using the term declare in the wrong way. what does it mean "exactly"?

I suppose it depends on your point of view. However, reread what I said, and you'll see that I already qualified the statement:

Quote:

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Originally Posted by Quzah

provided you've actually got the variable itself, and not just the value.

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You neglected to include that in your partial quote.

It depends how pedantic you want to be. In short, I'm right. Because, well, I've very seldom wrong when it comes to this sort of thing. But that's an aside.

In long...

Everything in C is passed by value. Functions either take values contained in variables, or they take values which are the addresses of other variables (or functions). If you say you pass a "single character", you mean one of two things:

1 - You are passing an integer value, which would fit in the allowede value for the type 'char'.
2 - You are passing the address of the 'char', and as such, are "passing a char"(acter).

If you have a function which wants a pointer to a character, it means one of two things:

1 - It's pretending to be a "string" function. "strings" are zero or more values as per 1 above, terminated by a nul character.
2 - It really wants you to pass it one char(acter), so that it can change the value in the actual char(acter) variable.

Functions which take values (and not address[es | values]) of variables, cannot change the contents of a variable passed to them. That is to say:

Code:

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char x;
...
foo( x );

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The variable x cannot be updated by foo(). It will remain unchanged, even if foo happens to do this:

Code:

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void foo( char bar )
{
    bar =! bar;
}

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x will not be changed. The function foo only gets the value of whatever was in x, not the actual variable.

So, were we actually to use the definition "pass [a] single character[s] to a function", this wouldn't qualify. It just gets a value. Not a variable. If you want to pass a variable so that it can be modified, you need to pass it via pointer. Thus, you copy its address to the function, so the function now can access the actual variable.

String functions take addresses of one character. (Because there's no such thing as one address representing more than one variable.) They treat that address as the start of a block of memory they're supposed to be able to play with, until they hit a value which equals the nul character.

Thus, they take one character as an argument. They move foreward in memory from it, until they find "HEY STOP!", then they do. Or should.

The point is, you can pass a single character to a string function. If someone wants to prove me wrong, go ahead. I need some entertainment.*


Quzah.


*Oh, and if you say "I meant the value of a character...", I automatically counter with: "In theory, the value 'a' could represent a spot in memory which just happens to point to an allocated block of memory my program has access to, which contains a single character. So while my compiler will warn about converting an integer to a pointer without a cast, it will be allowed, and could in fact not crash. Thus, we could in theory pass 'a' to a function, that just have the same value of an allocated 'char', containing the character 'a'."

Quote:

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Originally Posted by Abda92

i'm sorry i think i'm using the term declare in the wrong way. what does it mean "exactly"?

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I've written about forty posts explaining pointers, I'm sure you could find one if you wanted to. Anyway, what's one more...


Variables hold values. Pointers are variables that hold, as their values, addresses of a specific type*.

Code:

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char x;

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The variable x can hold the allowed ranges for type char. This is typically letters and numbers, punctuation, etc. However, you can ignore the fact that these values just happen to coincide with the ASCII (or your language setting of choice), and treat them as tiny integers. (Because that's really what they are.)

Now, pointers are used to "point at" (hence the name) other actual variables. They also hold values, but their value is the address of the type of variable to which they're allowed to point.

Code:

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char *p;

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The * means that p is a pointer. Read it right to left "p *(points to a) char. The value contained in p represents the address of a char, or NULL, which means it's not pointing at anything. Really, it means "This points at NOTHING."

That's important, because "uninitialized" and "at NOTHING" are two entirely different things.

Back on topic, to point at a variable, you have to be able to get its address. To do this, you use the & (address-of) operator.

Code:

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char x;
char *p;

p = &x;

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This means "take the address-of x and stick it in the variable p. In other words, "make p point to x".

To "get to" whatever something points at, we need to "dereference" the pointer. To do this, you use the * operator again.

Code:

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char x, y;
char *p;
...
p = &x;
...
y = *p;

---

This means "get the value stored in the variable that p is pointing at, and place that in y."

This is called 'dereferencing'.


Quzah.

*excluding void pointers. Void pointers can point at any normal type of object, and must be typecast correctly to be dereferenced. They cannot point to functions.

Quote:

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Originally Posted by Abda92

i'm sorry i think i'm using the term declare in the wrong way. what does it mean "exactly"?

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It is specifying the type of your variable. In script languages variables are usually untyped. In javascript or awk script, for example, you can just write

Code:

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some_variable = 7.6

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but in C, Java, Pascal and the like, variables are given a specific type so that the compiler will know how much memory to allocate for them and what sort of operations can or can't be applied (so that it can warn you if you appear to be doing something silly, like Quzar :-).

This is a declaration in C:

Code:

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double some_variable;

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Quote:

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Originally Posted by Abda92

so what is the difference between data types and an identifier? i'm sorry i really am confused:confused:

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Some C data types are:

char, int, float, double

An identifier is the arbitrary name you decide on for your variables:

p, limit, i, last_character

ok. thank you everyone. i'v read the thread again and was able to understand almost everything. let me summerize everything then please tell me if i'm right or wrong:
1. to declare a pointer you type:

Code:

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int(or char) *ptr;

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2. to dereference a pointer you type:

Code:

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foo(*ptr);

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3. a function that takes the value of a variable cannot modify it, you must pass it the address of the varible.
So is that all right?

is that all right?

Well the "foo( )" part is a hypothetical function call. Dereferencing just gives you whatever the value is in the type it points to. So if we had a pointer to an int, and we wanted to get the value of what it points at:

Code:

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int x = 10;
int y;
int *p;

p = &x; /* put the address of x in pointer p */
y = *p; /* dereference p, which gives us the value x contains, and put that in y */

---

Quzah.

ok thanks everyone i really appreciate you help