passing pointers to functions

Question

3.00/5 (2 votes)

See more:

Hi Guys,

I've been scouring the net but I can't quite get my head around how to copy a char array and pass the values to another function.

Say I have 3 arrays;

char buf1[50], buf2[50], buf3[50]

I am trying to copy and pass the data within each buffer to another function, but all this talk of pointers has got my head in a spin and even the word "function" has lost all meaning to me!

Would really appreciate the help.

Posted 27-Mar-11 3:10am

Member 7773377

Updated 27-Mar-11 5:30am

Dalek Dave

v2

Add a Solution

Comments

Dalek Dave 27-Mar-11 11:31am

Edited for Grammar and Spelling.

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

User 7429338 · Accepted Answer · 2011-03-27T03:30:00

Solution 1

To copy one buffer to another, you can do this;

for(int i=0;i<50;i++)
    buf2[i] = buf1[i];

To pass the buffer to a function you can use this;

void myfunction(char *buffer)
{
    // do something with buffer here, f.e. set all elements to 0
    for(int i=0;i<;50;i++)
        buffer[i] = 0;

}

void main(void)
{
    char buf1[50], buf2[50], buf3[50];

    // initialise all elements of each buffer to 0, using myfunction()
    myfunction(buf1);
    myfunction(buf2);
    myfunction(buf3);
}

Posted 27-Mar-11 3:30am

User 7429338

Comments

Dalek Dave 27-Mar-11 11:31am

Good Answer.

[no name] 27-Mar-11 11:33am

Thanks! Your vote gave me gold reputation ;)

Nemanja Trifunovic 27-Mar-11 17:06pm

It is not a good idea to hardcode the array size within the function. It is better to pass it as another parameter instead.

[no name] 28-Mar-11 2:51am

I agree, and there should probably be more meaningful names than myfunction() and buf1/buf2/buf3 in a real aplication.
To answer the question on how to pass an array to another function though, I think this code demonstrates what to do.

Nemanja Trifunovic 28-Mar-11 10:24am

It is not about style here. Because array decays into pointers when passed as function arguments, there is no way to determine the size of the array from within a function. Therefore, either the size needs to be passed as a separate parameter, or array needs to be terminated by some well-known value that the function will recognize.

Eddy Vluggen 28-Mar-11 3:16am

It's not a lecture on best-practices on coding, it's a valid and correct answer. Stay on-topic, please :)

Nemanja Trifunovic 28-Mar-11 10:26am

IMHO, it is an incomplete answer. OP wants to know how to pass a multidimensional array to a function. Passing just a pointer without the size(s) is not enough.

Eddy Vluggen 28-Mar-11 10:38am

I see your point; might have been a problem. Good to have a warning in advance - I mistook it for pedantry, like my own post :)

OriginalGriff 28-Mar-11 3:10am

Reason for my vote of 5: Good answer, and it puts you back to 4999 Authority...:laugh:

Andrew Brock · Accepted Answer · 2011-03-27T03:32:00

Surely there is a good explanation somewhere, but I will write my own.

In C/C++ (and pretty much every other language) an array is just a pointer to the first element in the array.
In C/C++ it also allocates memory on the stack, rather than you allocating memory with new, but that isn't really important to how they work.

So, we have 3 arrays (pointers) buf1, buf2 and buf3.
Each of these is just a pointer to a peice of memory, for this example:
buf1=0x1000
buf2=0x1100
buf3=0x1200

Then the contents of memory starting at 0x1000 is the elements of the array, say for example you have the code:

char buf1[50] = "Hello";
char buf2[50] = "World";
char buf3[50] = "Array";

Then the memory would look like this:

Offset 0x1000:
'H', 'e', 'l', 'l', 'o', '\0', (rest of 50 chars in undefined)
Offset 0x1100:
'W', 'o', 'r', 'l', 'd', '\0', (rest of 50 chars in undefined)
Offset 0x1200:
'A', 'r', 'r', 'a', 'y', '\0', (rest of 50 chars in undefined)

Where each character is 1 byte.

When you pass this pointer into a function, it just passes a copy of the memory location, for example:

void Print(char *str) {
	printf(str); //print variable <str> to the screen. <str> points to a bit of memory that has a string.
}

int main() {
	char buf1[50] = "Hello"; //Memory location 0x1000
	char buf2[50] = "World"; //Memory location 0x1100
	char buf3[50] = "Array"; //Memory location 0x1200
	Print(buf1); //This calls Print(0x1000)
}

Then what happens is the pointer to buf1 is passed to Print(), which is the memory location 0x1000.
This does not create a copy of the memory, and so if the function Print() changes anything in the memory, it will affect the string in both Print() and main().

If nothing is changed in Print(), in this example there isn't, then there is no need to copy the memory.
If you change something and you want this change to go back to the calling function main() (such as convert to upper case) you MUST NOT copy the memory (unless you want a copy in the original case as well).
If something is changed in Print() and you DON'T want this to change the array in main() then you need to copy the array, this can be done with another buffer or some allocated memory:

char buf1[50] = "Hello";
char buf1Copy[50];
strcpy(buf1Copy, buf1); //Copies the string (5 characters + 1 null terminating character '\0'). Only works for strings.
/* OR... */
memcpy(buf1Copy, buf1, sizeof(buf1)); //Copies all 50 bytes of buf1 into buf1Copy. Works for any array.