walking through an array of char pointers

[gaga]

I can't seem to get this to work:

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

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0') {
printf("%s\n",* np);
np++;
}

return 0;
}

after printing the 3 names, it prints garbage then sometimes seg faults.
I know i can do it easily using a for loop, but that's not what I am looking
for. I was under the impression names is actually:

[] --> "jack\0"
[] --> "jill\0"
[] --> "zack\0"
[] --> \0
If, that is so, shouldn't I be able to perform the above loop?

This works:

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

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);

return 0;
{

But, I need 3 loops for what I'd like to do in one.
I do not want to use the fact that I know there are n
strings in the array. I want to use the fact that each string is
null terminated then there is a final null terminator.

any suggestions would be appreciated.

- gaga

[Jens.Toerring]

gaga <zi*******@aol. com> wrote:

I can't seem to get this to work: #include <stdio.h>
#include <stdlib.h>
#include <string.h> int main()
{
char *names[3];
char **np; names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0') {
np hasn't been in initialized at this point. I guess you meant to
set
np = names;

before the start of the loop. But even then there's no reason why
the loop should stop after printing "zack" because afterwards np
would get incremented to point _after_ the third array element,
which probably is some random bit pattern which, when interpreted
as a char pointer, doesn't point to something where a '\0' charac-
ter would be stored. So you would need a fourth element of your
'names' array that would point to an empty string.
printf("%s\n",* np);
np++;
}

return 0;
}
You would need something like this instead to get it working:

int main( void )
{
char *names[ ] = { "jack", "jill", "zack", "" };
char **np = names;

while ( **np != '\0' )
printf( "%s\n", *np++ );
return EXIT_SUCCESS;
}
This works: #include <stdio.h>
#include <stdlib.h>
#include <string.h> int main()
{
char *names[3];
char **np; names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);

return 0;
{

If this version works for you it's just by accident - on my machine it
crashes immediately. And what's "\%s" and "*(*np)++" supposed to do?

Regards, Jens
--
\ Jens Thoms Toerring ___ Je***********@p hysik.fu-berlin.de
\______________ ____________ http://www.toerring.de

[CBFalconer]

gaga wrote:

I can't seem to get this to work:

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

int main()
{
char *names[3]; 4 /* need a final pointer */ char **np;
np = &names; /* initialize */ names[0] = "jack";
names[1] = "jill";
names[2] = "zack"; names[3] = ""; /* end marker */
while (**np != '\0') {
printf("%s\n",* np);
np++;
}

return 0;
}

after printing the 3 names, it prints garbage then sometimes seg
faults. I know i can do it easily using a for loop, but that's
not what I am looking for. I was under the impression names is

Try the commented changes/additions above. Simpler is:

char *names[] = {"jack", "jill", "zack", ""};
char* *np;

for (np = &names; **np; np++) printf("%s\n, *np);
--
fix (vb.): 1. to paper over, obscure, hide from public view; 2.
to work around, in a way that produces unintended consequences
that are worse than the original problem. Usage: "Windows ME
fixes many of the shortcomings of Windows 98 SE". - Hutchison

[Joe Wright]

CBFalconer wrote:

gaga wrote:

I can't seem to get this to work:

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

int main()
{
char *names[3];

4 /* need a final pointer */

char **np;

np = &names; /* initialize */

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

names[3] = ""; /* end marker */

while (**np != '\0') {
printf("%s\n",* np);
np++;
}

return 0;
}

after printing the 3 names, it prints garbage then sometimes seg
faults. I know i can do it easily using a for loop, but that's
not what I am looking for. I was under the impression names is

Try the commented changes/additions above. Simpler is:

char *names[] = {"jack", "jill", "zack", ""};
char* *np;

for (np = &names; **np; np++) printf("%s\n, *np);

char *names[] = {"jack", "jill", "zack", NULL };
char **np = names;
while (*np) printf("%s\n", *np++);

Note that names is an array of pointers to char. 'names' decays to a
pointer to the array's first element. A pointer to char. So the
assignment is 'np = names;', not 'np = &names;'.

--
Joe Wright mailto:jo****** **@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[Martin Ambuhl]

gaga wrote:

I can't seem to get this to work:

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

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0') {
printf("%s\n",* np);
np++;
}

return 0;
}

after printing the 3 names, it prints garbage then sometimes seg faults.
I know i can do it easily using a for loop, but that's not what I am looking
for. I was under the impression names is actually:

[] --> "jack\0"
[] --> "jill\0"
[] --> "zack\0"
[] --> \0

That would be true if you had declared an array of four pointers to char
(count your list) and set the last one to point to an empty string. You
obviously did not do that.
char *names[] = {"Jack", "Jill", "Zack", ""};

[tigervamp]

zi*******@aol.c om (gaga) wrote in message news:<56******* *************** ****@posting.go ogle.com>...

I can't seem to get this to work:

#include <stdio.h> #include <stdlib.h>
#include <string.h>
You don't need those last two.

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0') {
Here is your first problem. You are dereferencing a pointer (np)
which you haven't assigned to yet. I'm not sure what you are trying
to do here.
printf("%s\n",* np);
Don't know how this could possibly work...
np++;
}

return 0;
}

after printing the 3 names, it prints garbage then sometimes seg faults.
That's because you are trying to access memory pointed to by a pointer
that hasn't been initialized with a valid location.
I know i can do it easily using a for loop, but that's not what I am looking
for. I was under the impression names is actually:

[] --> "jack\0"
[] --> "jill\0"
[] --> "zack\0"
Good so far...
[] --> \0
Nope. "char *names[3];" declares an array of 3 pointers to char,
that's it. You may be getting confused with the '\0' that is
automatically added to string literals.

There are a few ways to do what you want to do. You can keep track of
how many elements are in the array and use a for loop to iterate over
them or you can do something like the following which uses a NULL
pointer as the last element in the array and is in the spirit of what
I think you were trying to do:

#include <stdio.h>

int main(void)
{
char *names[4];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";
names[3] = NULL;

np = names;

while (*np) {
printf("%s\n",* np);
np++;
}

return 0;
}

If, that is so, shouldn't I be able to perform the above loop?

This works:

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

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);

return 0;
{

There are more problems with this code than I care to comment on, get
rid of it.

Rob Gamble

[pete]

gaga wrote:

I can't seem to get this to work:

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

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0') {
printf("%s\n",* np);
np++;
}

return 0;
}

#include <stdio.h>

int main(void)
{
char *names[] = {"jack", "jill", "zack", NULL};
char **np;

for (np = names; *np != NULL; ++np) {
puts(*np);
}
return 0;
}

--
pete

[Joe Wright]

pete wrote:

gaga wrote:

I can't seem to get this to work:

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

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0') {
printf("%s\n",* np);
np++;
}

return 0;
}

#include <stdio.h>

int main(void)
{
char *names[] = {"jack", "jill", "zack", NULL};
char **np;

for (np = names; *np != NULL; ++np) {
puts(*np);
}
return 0;
}

Ok, printf is too complicated but for is too.

#include <stdio.h>

int main(void) {
char *names[] = {"jack", "jill", "zack", NULL};
char **np = names;
while (*np) puts(*np++);
return 0;
}

--
Joe Wright mailto:jo****** **@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[gaga]

ro**********@ho tmail.com (tigervamp) wrote in message news:<c9******* *************** ****@posting.go ogle.com>...

zi*******@aol.c om (gaga) wrote in message news:<56******* *************** ****@posting.go ogle.com>...

I can't seem to get this to work:

#include <stdio.h>

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

You don't need those last two.

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0') {

Here is your first problem. You are dereferencing a pointer (np)
which you haven't assigned to yet. I'm not sure what you are trying
to do here.

printf("%s\n",* np);

Don't know how this could possibly work...

np++;
}

return 0;
}

after printing the 3 names, it prints garbage then sometimes seg faults.

That's because you are trying to access memory pointed to by a pointer
that hasn't been initialized with a valid location.

I know i can do it easily using a for loop, but that's not what I am looking
for. I was under the impression names is actually:

[] --> "jack\0"
[] --> "jill\0"
[] --> "zack\0"

Good so far...

[] --> \0

Nope. "char *names[3];" declares an array of 3 pointers to char,
that's it. You may be getting confused with the '\0' that is
automatically added to string literals.

There are a few ways to do what you want to do. You can keep track of
how many elements are in the array and use a for loop to iterate over
them or you can do something like the following which uses a NULL
pointer as the last element in the array and is in the spirit of what
I think you were trying to do:

#include <stdio.h>

int main(void)
{
char *names[4];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";
names[3] = NULL;

np = names;

while (*np) {
printf("%s\n",* np);
np++;
}

return 0;
}

If, that is so, shouldn't I be able to perform the above loop?

This works:

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

int main()
{
char *names[3];
char **np;

names[0] = "jack";
names[1] = "jill";
names[2] = "zack";

while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);
np++;
while (**np != '\0')
printf("\%s\n", *(*np)++);

return 0;
{

There are more problems with this code than I care to comment on, get
rid of it.

Rob Gamble

Guys and gals....

you suggestions have been helpful, much appreciated, and correct.
But my blunder caused me to not get my point across.
I did not initialize **np, yes, but that was a copy paste mistake.

So, I guess where I'm really troubled is here:

All the solutions were right, but i'm gonna use Joe's here:

char *names[] = {"jack", "jill", "zack", NULL };
char **np = names;
while (*np) printf("%s\n", *np++);
Ok, that works, but, why is the last NULL needed in names?
Shouldn't that be done already? Why does this fail?

char *names[] = {"jack", "jill", "zack"};
char **np = names;
while (*np) printf("%s\n", *np++);
I guess that's what I'm driving at, sorry for the fck up earlier.

Please correct me if i am wrong, isn't this:

char c[] = "yes"

really:

"yes\0"
If that is true, shouldn't this:

char *names[] = {"jack", "jill", "zack"};

really be this:

"jack\0", "jill\0", "zack\0", \0

why isn't a final null terminator appened after an array of pointers?

Is it a case of, "that's just how it is"?

a final clarification would be most appreciated.

thanks again,
gaga

[CBFalconer]

Joe Wright wrote:

.... snip ...
char *names[] = {"jack", "jill", "zack", NULL };
char **np = names;
while (*np) printf("%s\n", *np++);

Note that names is an array of pointers to char. 'names' decays
to a pointer to the array's first element. A pointer to char. So
the assignment is 'np = names;', not 'np = &names;'.

I don't think so. names is locally declared, so the type being
used in the assignment is not subject to decay to a pointer. We
want a pointer to a char*, which is not the type of 'names',
because it is still an array. The value of "sizeof names" should
point this out. That is one reason I separate the *s in the
declaration:

char* *np = &names;

--
fix (vb.): 1. to paper over, obscure, hide from public view; 2.
to work around, in a way that produces unintended consequences
that are worse than the original problem. Usage: "Windows ME
fixes many of the shortcomings of Windows 98 SE". - Hutchison