pointer to array

**nitin1** · 07-26-2005

Dear Sir,
I have a probleme in a program in which I cant send the address of a pointer to an array ,like this,

char (*a)[10][10];
char *temp;
-----
a[0]=hh ------
a[1]=kk --------
temp=a[0];
a[0]=a[1];
a[1]=temp;

-------------
-------------

The output must be a[0]=kk
a[1]=hh

But there is an error in the last three steps,temp is not reciving the address of a[0],a[0] not reciving address of a[1].

I dont know how to locate the error,what is the problem with the pointer temp.
Plz remove my confiusion,
Thank you.

**itsme86** · 07-26-2005

What are hh and kk? Are those strings? Could you post the entire code segment?

Meanwhile, you can assign the address of a pointer to another pointer (a pointer to a pointer) and use it like so:

Code:

{
  char *ptr = "this is a string";
  char **pptr = &ptr;

  printf("The string is: %s\n", *pptr);
  printf("The first character of the string is: %c\n", (*pptr)[0]);
  printf("pptr: %p, *pptr: %p, &ptr: %p, ptr: %p\n",
    pptr, *pptr, &ptr, ptr);
}

Which should output:

The string is: this is a string
The first character of the string is: t
pptr: 0xbffff784, *pptr: 0x80484ac, &ptr: 0xbffff784, ptr: 0x80484ac

You can see that the address pptr points to is the address of the ptr pointer.

**nitin1** · 07-27-2005

To
itsme86,
cprogramming.
Sir,
First I will thank for my question.Plz try to understand me , i want a program which will swap the strings .

For this reason we need pointers & i take here a pointer to array.

here is the full code;

Code:

void main()
{

char (*a)[10][10];
char *temp;
int i;

    printf("enter the names");
for(i=0;i<=1;i++)
    scnaf("%s",a[i]);

temp=a[0];  /*assignig the add. of a[0] to temp*/
a[0]=a[1];   /*assiging the add. of a[1] to a[0]*/
a[1]=temp;   

(HERE is an error in above swapping )


pritnf("%s %s",a[0],a[1]);
}

_____________
out put must be,

enters the names nitin
sharma

sharma nitin
____________

This is my probleme why I cant send the address of a pointer to array((*a)[10][10]) to pointer of type char(*temp).

Thank you,
nitin.

**Salem** · 07-27-2005

Paste code you compiled, not what you 'remembered'
> scnaf("%s",a[i]);
> pritnf("%s %s",a[0],a[1]);
This is just a typo-fest

> char (*a)[10][10];
Start with char a[10][10];
And char temp[10];

Then the swapping would be
strcpy( temp, a[0] );
strcpy( a[0], a[1] );
strcpy( a[1], temp );

Oh, and main returns int, not void

**dwks** · 07-27-2005

i want a program which will swap the strings

Code:

int main(void) {
    char *s1, *s2, *p;

    /* initialize s1 and s2 */

    printf("\n%s\n%s\n", s1, s2);
    p = s1, s1 = s2, s2 = p;
    printf("\n%s\n%s\n", s1, s2);

    return 0;
}

**cbastard** · 07-27-2005

void main() wats the big deal in this?.it doest matter wat main returns.i know it returns 0 if execution was successful and else otherwise.

**confuted** · 07-27-2005

Exactly. It returns EXIT_SUCCESS if successful, otherwise it returns EXIT_FAILURE. It matters to people like me, who like to type commands like
./configure && make && sudo make install
on the Linux prompt. You'll have difficulty returning anything from a void main().

**quzah** · 07-27-2005

Originally Posted by cbastard

void main() wats the big deal in this?.it doest matter wat main returns.i know it returns 0 if execution was successful and else otherwise.

Ignore this poor attempt at trolling. If you're really that ignorant, just read the FAQ. Nothing to see here, move along.

Quzah.

**cbastard** · 07-27-2005

Originally Posted by nitin1

To
itsme86,
cprogramming.
Sir,
First I will thank for my question.Plz try to understand me , i want a program which will swap the strings .

For this reason we need pointers & i take here a pointer to array.

here is the full code;

Code:

void main()
{

char (*a)[10][10];
char *temp;
int i;

    printf("enter the names");
for(i=0;i<=1;i++)
    scnaf("%s",a[i]);

temp=a[0];  /*assignig the add. of a[0] to temp*/
a[0]=a[1];   /*assiging the add. of a[1] to a[0]*/
a[1]=temp;   

(HERE is an error in above swapping )


pritnf("%s %s",a[0],a[1]);
}

_____________
out put must be,

enters the names nitin
sharma

sharma nitin
____________

This is my probleme why I cant send the address of a pointer to array((*a)[10][10]) to pointer of type char(*temp).

Thank you,
nitin.

thats one of the possible answer.ur statements a[0]=a[1] is wrong i think.thats the difference between pointer and array.u can not use array pointer to left side of assignment operator.
int main()
{

char a[10][10];
char temp[10];
int i;

printf("enter the names");
for(i=0;i<=1;i++)
scanf("%s",a[i]);

strcpy(temp,a[0]); /*assignig the add. of a[0] to temp*/
strcpy(a[0],a[1]); /*assiging the add. of a[1] to a[0]*/
strcpy(a[1],temp);

printf("%s %s",a[0],a[1]);

return 0;
}

**dwks** · 07-27-2005

Use code tags.

What's this for?

Code:

for(i=0;i<=1;i++)

[edit]
Never mind, I see it's supposed to be i<=10.
[/edit]

**dwks** · 07-27-2005

u can not use array pointer to left side of assignment operator

Yeah, you should use a for loop or memset or strcpy or something else.

**dwks** · 07-27-2005

Actually, I don't think you're supposed to change where an array points to. That is,

Code:

char *p, *q;
/* ... */
p = q;

is valid, but

Code:

char p[100], *q;
/* ... */
p = q;

is not.

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