C / C++ / MFC

passing array of pointers

15-Nov-06 15:29

I assume the dialogs are modeless ? If so, you need to call a method, instructing them to resize. If not, you should set them up to pass in a size when you create them.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 15:18

15-Nov-06 15:18

Hello all,
I cannot access names[0] and names[1] inside function B. It is null if i try to access them. Can anyone help ?

funcA{<br />
char * Names[] = {};<br />
int next = 0;<br />
char * list = "Abc";<br />
<br />
<br />
Names[ next ] = malloc( sizeof( char )*( strlen( list ) + 1 ) );<br />
memcpy( Names[ next ],list,sizeof( char )*( strlen( list ) + 1 ) );<br />
<br />
list = "cdefg";<br />
Names[ next+1 ] = malloc( sizeof( char )*( strlen( list ) + 1 ) );<br />
memcpy( Names[ next+1 ],list,sizeof( char )*( strlen( list ) + 1 ) );<br />
<br />
printf("vsp list got is %s\n", Names[0] );<br />
printf("vsp list got is %s\n", Names[1] );<br />
<br />
funcB(....,...,(const char**)&Names);<br />
}<br />
<br />
funcB(...,...,const char ** names)<br />
{<br />
....<br />
//the actual values names[0] and names[1] are to be accessed and used but they are null!!!???<br />
...<br />
//free names[0] and names[1]<br />
free( (char*)names[0]);<br />
free( (char*)names[1]);<br />
<br />
}

Christian Graus15-Nov-06 15:26

15-Nov-06 15:26

This is C, not C++ ?

I believe you want to cast Names to char **, and not the address of Names. I think an array is a pointer by definition. You could also declare Names as char **.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 15:37

15-Nov-06 15:37

This is C. this is my first small exercise in pointers and arrays.

a.)Where do you say I have to cast Names to char**?

I should be able to access all the names inside function B. How should i do it?

b.)If i declare as char ** Names, then how do I do malloc on Names[0]?

It will be great help for me to understand better if you could show/make changes in the code snippet.

Thanks for your time.

Christian Graus15-Nov-06 15:41

15-Nov-06 15:41

In C++, I would do this:

char ** names = new char * [2];
names[0] = new char[len1];
names[2] = new char[len2];

Where len1 and len2 are the lengths of my strings. I guess you can use malloc in the same way.

I'm saying that when you pass &Names, I think it's a char ***, because a char * [] is the same as a char **. At least, I think so.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 15:50

15-Nov-06 15:50

char ** names = new char *[2];

but I do not know if the names would be just 2 ...say if it is to be any number of names. then how do we do it?

Christian Graus15-Nov-06 16:02

15-Nov-06 16:02

I see what you'e doing. You can't dynamically add to an array. You can put a variable in place of the 2, but you can't add to the list, to do that you need to write a dynamically allocated array ( which typically keeps some space in reserve, and then reallocates itself to a new memory block when more is needed ).

C++ has classes to do this, I don't believe there is anything in C.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 16:49

15-Nov-06 16:49

thanks for that.
If Say, i know the dimension of the array.

int count  =4;<br />
int i = 0;<br />
char * Names[count];<br />
<br />
//to initialise<br />
for(i=0; i < count; i++)<br />
{<br />
Names[i] = NULL;<br />
}<br />

and then do the rest. Is the initialisation correct?
is what i have done equivalent to doing
char *Names[4] = {NULL,NULL,NULL,NULL};

Christian Graus15-Nov-06 16:51

15-Nov-06 16:51

Yes, I believe so. Again, I think your core problem is solved by passing Names instead of &Names.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 17:18

15-Nov-06 17:18

modified. ....
thanks.
i have another issue:

const char * str1 = "abcdef";<br />
const char * str2;<br />
when i try to do:<br />
strcpy(str2,str1);<br />

there is warning message.-- passing arg 1 of `strcpy' discards qualifiers from pointer target type
but when i try to run the code (using gcc),
the code crashes at strcpy()
why? how to fix this warning?

Christian Graus15-Nov-06 17:21

15-Nov-06 17:21

strcpy copies the value of a string, I don't believe it actually allocates the memory for you. You need to make str2 point to valid memory first.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 17:23

15-Nov-06 17:23

sorry i have modified my previous post. i do get a warning.
passing arg 1 of `strcpy' discards qualifiers from pointer target type

Christian Graus15-Nov-06 17:30

15-Nov-06 17:30

probably the const Smile | :)

Christian Graus - C++ MVP

thathvamsi15-Nov-06 18:55

15-Nov-06 18:55

hi...I tried doing the following:

const char* str1 = "abcd";
char* str2;
strcpy(str2,str1);

There is no compilation error or warning. but the program crahses at this stcpy line. Why would it?

Can I do this?
str2= str1;
will this work? what happens when i dothis?

Christian Graus15-Nov-06 19:12

15-Nov-06 19:12

*sigh* please read my answer from before. str2 should NEVER be declared like this, set all pointers to NULL until you give them a valid value. str2 is not valid, and it's pointing at junk, hence the error.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 20:06

15-Nov-06 20:06

thanks. fixed it.
i need to split a string in to tokens using delimiter but without using the strtok( ) or the strtok_r functions.
ie., const char * str1 = "apple+mango+grapes+orange";
i need to spli them into
apple
mango
grapes
orange
the delimter here is "+".
But is there a way (code snippet) that would do the above without using the strtok functions?

Christian Graus15-Nov-06 20:10

15-Nov-06 20:10

Yes, you can do what strtok does, make a temporary pointer, and keep pointing to the place past the last delimited ( first replacing the delimiters with NULL ).

Christian Graus - C++ MVP

thathvamsi15-Nov-06 21:45

15-Nov-06 21:45

Thanks very much for your help. It has been a good learning experience.
Back to my sqaure1 problem,,,,In the following code, the names are null when i try to access them inside funcB. why?
Should I do malloc for Names? if so, how?

funcA{<br />
array = 4;<br />
char * Names[array];<br />
int next = 0;<br />
int i;<br />
<br />
//initialise <br />
for(i=0;i<array+1;i++)<br />
{<br />
Names[i]=NULL;<br />
}<br />
<br />
char * list = "Abc";<br />
Names[ next ] = malloc( sizeof( char )*( strlen( list ) + 1 ) );<br />
memcpy( Names[ next ],list,sizeof( char )*( strlen( list ) + 1 ) );<br />
<br />
list = "cdefg";<br />
Names[ next+1 ] = malloc( sizeof( char )*( strlen( list ) + 1 ) );<br />
memcpy( Names[ next+1 ],list,sizeof( char )*( strlen( list ) + 1 ) );<br />
<br />
printf("vsp list got is %s\n", Names[0] );<br />
printf("vsp list got is %s\n", Names[1] );<br />
<br />
funcB(....,...,(const char**)&Names);<br />
}<br />
<br />
funcB(...,...,const char ** names)<br />
{<br />
....<br />
//the actual values names[0] and names[1] are to be accessed and used but they are null!!!???<br />
...<br />
//free names[0] and names[1]<br />
free( (char*)names[0]);<br />
free( (char*)names[1]);<br />
<br />
}

Christian Graus15-Nov-06 21:56

15-Nov-06 21:56

thathvamsi wrote:
funcB(....,...,(const char**)&Names);

I've told you at least twice to do this:

funcB(....,...,(const char**)Names);

because your array IS a char **, you're passing a char ***.

Christian Graus - C++ MVP

thathvamsi15-Nov-06 22:12

15-Nov-06 22:12

I did and it still gives null only.

Christian Graus15-Nov-06 22:30

15-Nov-06 22:30

To be honest, I tried to do a sample project for you, and found that VC will not compile most of your code. What you've posted, does not compile, and cannot compile. Even fixing the typos, VC will not compile it, another ( very non conformant ) compiler might, but VC will not.

Try creating a char **, like I said in the first place.

Christian Graus - C++ MVP

Christian Graus15-Nov-06 22:35

15-Nov-06 22:35

This actually works just fine. Obviously, all funcB does is free the names, but they are correctly being passed through.

#include <string.h>
#include <malloc.h>

void funcB(const char ** names)
{
//the actual values names[0] and names[1] are to be accessed and used but they are null!!!???


//free names[0] and names[1]
free( (char*)names[0]);
free( (char*)names[1]);

}

void funcA()
{
int array = 4;
char ** Names = new char *[array];
int next = 0;
int i;

//initialise
for(i=0;i < array; ++ i)
{
    Names[i]=NULL;
}

char * list = "Abc";
Names[ next ] = (char*) malloc( sizeof( char )*( strlen( list ) + 1 ) );
memcpy( Names[ next ],list,sizeof( char )*( strlen( list ) + 1 ) );

++next;

list = "cdefg";
Names[ next ] = (char*) malloc( sizeof( char )*( strlen( list ) + 1 ) );
memcpy( Names[ next ],list,sizeof( char )*( strlen( list ) + 1 ) );

printf("vsp list got is %s\n", Names[0] );
printf("vsp list got is %s\n", Names[1] );

funcB((const char**)Names);
}

int main(int argc, char* argv[])
{
    printf("Hello World!\n");
    funcA();
    return 0;
}

Christian Graus - C++ MVP

thathvamsi15-Nov-06 23:08

15-Nov-06 23:08

Hi...
Thanks for your time and help. I fixed it and Iam using char** Names but obviously, there is a bug with teh gcc compiler for it allowed me to do char*Names[array];
and returned no error or warnings.

But the program works now.
Thanks again. You have been a great help to me in understanding pointers and arrays and also gcc.

Christian Graus15-Nov-06 23:11

15-Nov-06 23:11

No worries. If it accepted 'new', then it's compiling your code as C++.

I'm glad I helped. You'd do well to buy a book, or at least read up on the web on this stuff, to solidify the stuff we've worked on.

Christian Graus - C++ MVP

Christian Graus15-Nov-06 23:09