Access violation in Dev-cpp

[niyasc]

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
int main()
{
    int i=0,j=0,n=0;
    char *a,**b;
    a=(char *)malloc(sizeof(char)*50);
    printf("Enter your text\n");
    gets(a);
    while(a[i]!='\0')
    {
                     
                     *(b+n)=(char *)malloc(sizeof(char)*20);  //error
                     for(j=0;a[i]!=' ';i++,j++)
                     {
                                     *(*(b+n)+j)=*(a+i);
                     }
                     i++;
                     n++;
    }
    
    printf("Words in given text are \n");
    for(i=0;i<n;i++)
    {
                    puts(*(b+i));
                    printf("\n");
    }
    
    printf("Possibile combinations are \n\n");
    for(i=0;i<n;i++)
    {
                    for(j=0;j<n;j++)
                    {
                                    if(j!=i)
                                    {
                                            puts(b[i]);
                                            printf(" ");
                                            puts(b[j]);
                                            printf("\n");
                                    }
                    }
    }
    getch();
}

Above code is for separating words from a given line of text and to produce all possible two word combinations. When I execute this program in dev-cpp

*(b+n)=(char *)malloc(sizeof(char)*20);
produces an access violation.Actually what is wrong here. When i execute same program in turbo c, it does not produce expected output. Can you help me?

[Narue]

Your b pointer is completely uninitialized.

[niyasc]

Your b pointer is completely uninitialized.

I'm using this code to initialize b pointer
*(b+n)=(char *)malloc(sizeof(char)*20);

what is wrong with this initialization?

[cse.avinash]

*(b+n) equivalent to b[n];
*(*(b+n)+j) equivalent to b[n][j];

**b will store the address the address of any pointer or an array.
I am little confused with your code...!
Can you please explain me what all these thing means..??

[Narue]

I'm using this code to initialize b pointer
*(b+n)=(char *)malloc(sizeof(char)*20);

No, that looks more like you're assuming b is already initialized, otherwise you wouldn't be derefencing it.

what is wrong with this initialization?

It's an initialization of b[n] , not b .

[niyasc]

No, that looks more like you're assuming b is already initialized, otherwise you wouldn't be derefencing it.

It's an initialization of b[n] , not b .

Do we have to initialize b and b[n]separately?

[niyasc]

Can anyone give me the proper code for above program?

[D33wakar]

I'm using this code to initialize b pointer
*(b+n)=(char *)malloc(sizeof(char)*20);

That's not called initializing ,it's called dereferencing of b.Dereferencing a pointer always involves the value present at the address pointed to by that pointer.
for e.g.
If there's a pointer 'ptr' to char

char letter1='A'];
char *ptr;

Then you initialize it like this

ptr=&letter1;

and dereference like this:

char letter2=*(ptr);//dereferencing ptr
printf("%c\n",letter2);

Do we have to initialize b and b[n]separately?

Here b is a pointer to a pointer to a char.So b should contain an address of a pointer to a char.
for e.g.

char *letters1="hello",**ptr;
ptr=&letters1;//initializing ptr(letters1 is itself a pointer to a string of char)
char *letters2=*(ptr);//dereferencing ptr
printf("%c\n",*(letters2);//points to the first char of the string "hello"
printf("%c\n",*(letters2+3));//points to the fourth(array's index starts from 0) char of the string "hello"

Now, do you realize what you are missing from the code you posted above?

[Narue]

Do we have to initialize b and b[n]separately?

They're two separate pointers, so yes:

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

int main(void)
{
    char **b = malloc(10 * sizeof *b);
    
    if (b != NULL) {
        int i, j, n = 0;
        char a[50];
        
        printf("Enter your text: ");
        fflush(stdout);
        
        if (fgets(a, sizeof a, stdin) != NULL) {
            for (i = 0; a[i] != '\0'; n++) {
                b[n] = malloc(20);
                
                if (b[n] == NULL)
                    break;
                
                for (j = 0; a[i] != ' ' && a[i] != '\0'; i++, j++)
                    b[n][j] = a[i];
                    
                b[n][j] = '\0';
                
                if (a[i] != '\0')
                    ++i;
            }

            printf("Words in given text are:\n");
            
            for (i = 0; i < n; i++)
                printf(">%s<\n", b[i]);
        }
    }

    return 0;
}

[niyasc]

Thanks for everyone
Here is my final code

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
int main()
{
    char **b,*a;
    int i,j,n=0;
    a=(char *)malloc(100);
    b=(char **)malloc(sizeof(*b)*20);
    printf("Enter your text\n");
    gets(a);
    for(i=0;a[i]!='\0';n++)
    {
                           b[n]=(char *)malloc(20);
                           for(j=0;a[i]!=' '&&a[i]!='\0';i++,j++)
                           {
                                           b[n][j]=a[i];
                           }
                           b[n][j]='\0';
                           if(a[i]!='\0')
                           {
                                         i++;
                           }
    }
    printf("Words in given text are \n");
    for(i=0;i<n;i++)
    {
                    printf("%s\n",b[i]);
    }
    printf("Possible combinations are \n");
    for(i=0;i<n;i++)
    {
                    for(j=0;j<n;j++)
                    {
                                    if(i!=j)
                                    {
                                            printf("\n%s %s",b[i],b[j]);
                                    }
                    }
    }
    getch();
}