Thread: "invalid use of void expression" error

Welcome to the forum, Smith283! (void)

Thnx laserlight for your reply but i didn't get it, you need to explain more please as i'm a beginer in C.
And actually the problem is in fetching the value inside ptr[i][j] i.e smith->ptr[i][j] . I have allocated the space for void type (size is 5) and now if i want to write into it say a string(characters), i have to just give a starting address of allocated space(&ptr[0][0]). It will successfully write in the allocated space. But if i want to fetch each character from it using "smith->ptr[i][j]" , it is giving an error "invalid use of void expression".

So did you get my actuall question?????

Thnx, hope you help.

The compiler complains about this line

Code:

smith->ptr[i] = malloc(sizeof(void) * 5);

You cannot get the size of void. void represents an unknown type and that has no size.
if you want to allocate 5 bytes (as stated in the previous post ) then use

Code:

smith->ptr[i] = malloc(sizeof(char) * 5);

because sizeof(char) is guaranteed to be 1.
or simply

Code:

smith->ptr[i] = malloc( 5 );

This loops don't make any sense to me

Code:

for(i=0;i<4;i++)
    {
        for(j=0;j<5;j++)
        {
            printf("address of ptr[%d][%d] is %p and the value inside it is %d \n",i,j,&smith->ptr[i][j],(int)smith->ptr[i][j]);
        }
    }
}

It seems that you are trying to print 5 ints but there is only space for 5 uninitialized chars

NOTE:
somehow I have the feeling that you wanted to declare struct me as

Code:

struct me {
    int ** ptr;
};

to have a 2 dim array of ints.

Kurt

Your mechanic wants to know what model and year of car you will be bringing in for service, later today.

You keep saying "a vehicle".

Your mechanic is not able to use that kind of non-specific information.

One question for all of you, if i comment the following lines as shown in the code below, it would work or not????

Code:

#include<stdio.h>
#include<stdlib.h>
 
struct me
{
        void **ptr;
 
}*smith;
 
 
int main()
{
        int i,j;
 
        smith = malloc(sizeof(struct me));
        smith->ptr=malloc(sizeof(void *) * 4);
 
        for(i=0;i<4;i++)
        {
                smith->ptr[i] = malloc(sizeof(void) * 5);
        }
 
        for(i=0;i<4;i++)
        {
                printf("address of ptr[%d] is %p and the value inside it is %p \n",i,&smith->ptr[i],smith->ptr[i]);
        }
 
 
/*        for(i=0;i<4;i++)
        {
                for(j=0;j<5;j++)
                {
                        printf("address of ptr[%d][%d] is %p and the value inside it is %d \n",i,j,&smith->ptr[i][j],(int)smith->ptr[i][j]);
                }
        }
*/

return 0;
}

Answer yes or no?? Please......

Extremly sorry mate, i should have told you all in the begining that i'm using GCC compiler (fedora 14, linux).
And iMalc, i think the program in the above lastest post of mine will work fine (without errors). The problem here is not the "sizeof(void)" bt error comes from the printf line that i've commented.
And it would allocate the memory for void type succcessfully:-

Code:

for(i=0;i<4;i++)        {

                smith->ptr[i] = malloc(sizeof(void) * 5);

        }

The error "invalid use of void expression" which the compiler is giving is for this line:-

Code:

printf("address of ptr[%d][%d] is %p and the value inside it is %d \n",i,j,&smith->ptr[i][j],(int)smith->ptr[i][j]);

If i try to print "sizeof(void)" it wont work, but how it is allocating memory "void" type in this line:-

Code:

    smith->ptr[i] = malloc(sizeof(void) * 5);

I need a perfect reason for this.

Also i've done writting into void type allocated space, if you people are right then how this is working:-

Code:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

struct me
{
        void **ptr;

}*smith;


int main()
{
        int i,j;
        char buff[5]="hello";

        smith = malloc(sizeof(struct me));
        smith->ptr=malloc(sizeof(void *) * 4);

        for(i=0;i<4;i++)
        {
                smith->ptr[i] = malloc(sizeof(void) * 5);
                memset(smith->ptr[i],'\0',sizeof(void) * 5);
        }

        for(i=0;i<4;i++)
        {
                printf("address of ptr[%d] is %p and the value inside it is %p \n",i,&smith->ptr[i],smith->ptr[i]);
        }

        strncpy(smith->ptr[0],buff,5);
//              or
//      strcpy(&smith->ptr[0][0],buff);


        printf("the string is : %s\n",smith->ptr[0]);
//              or
//      printf("the string is : %s\n",&smith->ptr[0][0]);


/*      for(i=0;i<1;i++)
        {
                for(j=0;j<5;j++)
                {
                        printf("address of ptr[%d][%d] is %p and the value inside it is %c \n",i,j,&smith->ptr[i][j],(char )smith->ptr[i][j]);    ****Compiler give error on this line not on the sizeof(void)****
                }
        }
*/

return 0;
}

Thnk You

smith

Does in linux "void" size is 1byte????

The problem here is not the "sizeof(void)" bt error comes from the printf line that i've commented.

Like so many newbies who think they know what they are doing you are only looking for confirmation of a self-inflicted misunderstanding.

I need a perfect reason for this.

You've been given the reason. You rejected that reason.

You want a "perfect" reason for this that meshes with your poor understanding of C.

Also i've done writting into void type allocated space, if you people are right then how this is working:-

It isn't; it doesn't.

"GCC" is letting you do this without complaining because an extension relevant to the history of C; the compiler doesn't care if you use this extension incorrectly by default.

Add "-pedantic" to your command line and the compiler will start complaining.

Soma

I have'nt rejected anything, the problem is i'm not getting it

sizeof(void) is invalid to use, i get it.

but would it allocate memory for void type or not in this line???????????????

"smith->ptr[i] = malloc(sizeof(void) * 5);"

If "yes" :This is your answer:-

" "GCC" is letting you do this without complaining because an extension relevant to the history of C; the compiler doesn't care if you use this extension incorrectly by default.

Add "-pedantic" to your command line and the compiler will start complaining. "

if "no" :then your answer is:-
" You've been given the reason. You rejected that reason. "
????
????