Thread: Pass a string to a function

Hi Guys,

I got stuck im my last thread so I have tried to simplify things to see where im going wrong,

In class we just covered basic C functions, I can pass an 'int' back and forth no problems but im getting stuck passing a "string".

This is not class work just me trying to get my head round learning basic C.

I dont know about pointers as yet this has not been covered and I dont think it will be at the level we are at so im trying to do this without pointers.

I want to take user input in the main() function and pass it to another function, manipulate the string in some way(in this case just add ABC to the string passed) and pass it back to the main() function and print it out with the new string3 added to the end of what the user input.

Any help is appreciated.
regards
c

Im using Borland 4.5 C++ Complier on WinXP, and its giving me the error:

Cannot convert 'char *' to 'char' in function main()

Code:

#include<stdio.h>
#include<conio.h>
#include<string.h>
#include <ctype.h>

/*Functions Prototype */
void check_str(char);


/*The main program*/

void main(void)
{
char string1[40];
char answer[40];


  printf("\n\n\t\t   Enter string1\n\n");
  printf("Please enter a string \n");
  gets(string1);


  strcpy(answer,check_str(string1));
  printf("answer = %s",answer);

}


check_str(char string2)
{
char string2[40];
char string3[40];
string3[4]={'a','b','c','\0'};


    strcat(string2,string3);
   return char string2[40];
}

it trips on this line of code

Code:

strcpy(answer,check_str(string1));

>>void main(void)
Always use int main()

>>void check_str(char);
As you defined this function, it should retun a string, and it gets a parameter that is a string, so it should be char* check_str(char *)

Code:

check_str(char string2)
{
char string2[40]; /*you don't need this*/

As your function gets string2 as parameter, you don't need to redeclare it inside the function.

Code:

char string3[40];
string3[4]={'a','b','c','\0'};

char string3[40] means a char array of 40 elements; string3[4] is the 5th element in the array, it can be a single char...
If you want to define a string, it should be more like

Code:

char string3[]="abc";

And you don't need to include <conio.h>

Got it guys,

Thanks a bunch.

I changed the return in my function to

Code:

return string2;

I realy need to figure out how to declare strings properly I was realy jumbled up there.

Thanks again
C

In function char *check_str(char *string2):

Code:

char *check_str(char *string2)
{
char string3[]="abc";

strcat(string2,string3);
return char string2[40];
}

don't return char string2[40], just return string2; - the variable has been already declared; you have to specify only the name of the variable you want to return.

And to the end of int main() add return 0;.
-------------------------------------
A little late, I see...

Hey Guys,

Just an after thought, Its nice having the program run but im not too sure what the new bits I add have realy done.

Where I have added char * what is that actually saying or doing, does this just mean the char is a string or is the * into pointers??

I looked through the C ref guide at:

http://www.infosys.utas.edu.au/info/....html#Contents

But never came up with much.

Thanks again
C

Originally Posted by Scribbler

char *<variable> is read from right to left as variable is a pointer to a type char. In C, strings are simply char arrays. So basically what your doing is passing the address of the first char of the array.

Additionally, since you didn't modify the value passed to the function. You could redefine the function to be of type Void, since you really didn't change the address of the char array (only the contents of the array) and therefore there's no need to return a value back.

Hey scribbler

<< where you say "So basically what your doing is passing the address of the first char of the array."

That is where the code says

Code:

char *check_reg(char *reg)

in the (char *reg) of the code yea?

And when this is ran it moves to the next address untill it hits an address that holds a null terminator??


What about the code that says:

Code:

/*Functions Prototypes*/
char * check_reg(char*);

What am I saying here is this still read right to left?


Thanks again
C