passing character to a function

What I basically have is:

Code:

---

int main()
{
  char * myText;
  myText = "blah";
  function(myText);
 
  return 0;
}


function(char * text)
{
  //do stuff
}

---

what I want to be able to do is pass a single character to my function, however when do this:

Code:

---

function(myText[1]);

---

It doesn't work and I get compiler errors.

What I want is the way to pass a single character to this function? Note that I still want to be able to pass whole arrays to this function.

Any help would be appreciated :)

Do this:

function(&myText[1]);

Single character = char.
Array of characters = char*.
Just remember that you cannot modify a string literal, thus the code:

Code:

---

  char * myText;
  myText = "blah";

---

...should be...

Code:

---

  const char * myText = "blah";

---

That will prevent mistakes.

Of course, you can pass a single character to a function. However, your function prototype requires a pointer to character, and as you are probably aware, a pointer to a character, and a character are not the same thing - just like the address where I live is not the same as my person.

Do you want to pass a pointer, or just a single character?

--
Mats

function(&myText[1]);
would work, but it doesn't help you figure out whether you meant 1 character or a string.

Both are the same as far as function is concerned.

Quote:

---

Originally Posted by matsp

Of course, you can pass a single character to a function. However, your function prototype requires a pointer to character, and as you are probably aware, a pointer to a character, and a character are not the same thing - just like the address where I live is not the same as my person.

Do you want to pass a pointer, or just a single character?

--
Mats

---

Sorry, I want to pass the pointer

So like Salem or Dino says, &myText[x] will pass the address of character x.

--
Mats

If I use &myText[1] It takes the address of the character, but all subsequent characters are also included..

so in this example, mytext = lah instead of l

What is your current code?

Quote:

---

Originally Posted by RedWind

If I use &myText[1] It takes the address of the character, but all subsequent characters are also included..

so in this example, mytext = lah instead of l

---

You fail to understand.
It takes the address to the character ONLY.
However, as it is a pointer (and a pointer is a variable), it can be incremented to the next character in memory.
In C, all strings are passed as char* and the memory is traversed byte by byte until the '\0' character is found. This gives the "illusion" that when you pass the char*, you get all subsequent characters, as well, but this is not correct.

Quote:

---

Originally Posted by Elysia

You fail to understand.
It takes the address to the character ONLY.
However, as it is a pointer (and a pointer is a variable), it can be incremented to the next character in memory.
In C, all strings are passed as char* and the memory is traversed byte by byte until the '\0' character is found. This gives the "illusion" that when you pass the char*, you get all subsequent characters, as well, but this is not correct.

---

ah of course, that makes sense. so all I have to do is set the next byte to '\0' and I've solved my 'problem'.

Indeed! And thus you've created a string with only 1 character!
That sounds silly, but it is a solution. Good job figuring it out, though. I commend you for that :)