char **argv vs. char* argv[]

[Bret]

I'm curious why char** argv is acceptable in the main() declaration.

In the comp.lang.c FAQ (question 6.18) it says that pointers to
pointers and pointers to an array are not interchangable. However the
declaration:

int main(int argc, char** argv)

is common.

How does that work out? Shouldn't the compiler complain if you're not
doing:

int main(int argc, char* argv[])

Thanks,
Bret

 

[Joona I Palaste]

I'm curious why char** argv is acceptable in the main() declaration.

In the comp.lang.c FAQ (question 6.18) it says that pointers to
pointers and pointers to an array are not interchangable. However the
declaration:

int main(int argc, char** argv)

is common.

How does that work out? Shouldn't the compiler complain if you're not
doing:

int main(int argc, char* argv[])

This is because in function parameter declarations, char **foo and
char *foo[] are the same thing. Not just in main(), but in every other
function too. What's more, int **foo and int *foo[] are also the same.
So are int ***foo and int **foo[].
In fact, in general, if T is a type, then T *foo and T foo[] are the
same thing in function parameter declarations. (This goes ONLY one
level deep - T **foo and T foo[][] are a different matter entirely!)
This is all explained by Chris Torek's THE RULE. If used as a value,
or as a function parameter, an array decays into a pointer into its
first element. Therefore you can't really pass arrays into functions
at all. You can pass pointers to arrays, or structures containing
arrays, but not arrays themselves.

--
/-- Joona Palaste ([email protected]) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"Roses are red, violets are blue, I'm a schitzophrenic and so am I."
- Bob Wiley

 

[Brett Frankenberger]

I'm curious why char** argv is acceptable in the main() declaration.

In the comp.lang.c FAQ (question 6.18) it says that pointers to
pointers and pointers to an array are not interchangable. However the
declaration:

int main(int argc, char** argv)

is common.

char **argv declares argv as a pointer to a pointer to a character
char *argv[] declares argv as am array of pointers to a character

There are no pointers to arrays in either case. So whether or not
pointers to arrays are interchangable with pointers to pointers is
irrelevant to whether "char **argv" and "char *argv[]" are both
allowable as the second parameter to main().

The relevant question you would want to ask is whether arrays and
pointers are interchangable. They aren't in the general case, in the
formal parameter list of a function, they more-or-less are with respect
to the outer-most nesting layer.

-- Brett

 

[E. Robert Tisdale]

I'm curious why char** argv is acceptable in the main() declaration.

In the comp.lang.c FAQ (question 6.18) it says that
pointers to pointers and pointers to an array are not interchangeable.
However the declaration:

int main(int argc, char** argv)

is common.

How does that work out?
Shouldn't the compiler complain if you're not doing:

int main(int argc, char* argv[])

It probably should complain.

> cat f.c

int f(char* s[]) {
char** p = s;
}

> gcc -Wall -std=c99 -pedantic -c f.c

f.c: In function `f':
f.c:2: warning: unused variable `p'
f.c:3: warning: control reaches end of non-void function

C performs an "implicit conversion"
from an array to a pointer to its first element
and from a pointer to an array.

I can write

> cat f.c

int f(char* s[]) {
char** p = s;
return f(p);
}

> gcc -Wall -std=c99 -pedantic -c f.c

and my compiler will not complain.

 

[Irrwahn Grausewitz]

Shouldn't the compiler complain if you're not doing:

int main(int argc, char* argv[])

It probably should complain.
Why?

cat f.c

int f(char* s[]) {
char** p = s;
}

gcc -Wall -std=c99 -pedantic -c f.c

f.c: In function `f':
f.c:2: warning: unused variable `p'
f.c:3: warning: control reaches end of non-void function

Your compiler complains about you having declared an automatic variable
without using it and to not return a value from a function that is
declared as returning int. This has no relation to the given problem.

C performs an "implicit conversion"
from an array to a pointer to its first element

Right, when used as a value or formal parameter.

and from a pointer to an array.

Huh?
ITYM one can apply the '[]' operator to a pointer like
one would do with an array, so that:

int i = 21;
int a[42];
int *p = a;

leads to:

p == a;

I can write

cat f.c

int f(char* s[]) {
char** p = s;
return f(p);
}

gcc -Wall -std=c99 -pedantic -c f.c

and my compiler will not complain.

Yes, because this time you use p *and* you (formally) return a
value while in fact f() won't return anything at all as it will
crash your program when called because of infinite recursion.

Irrwahn

 

[E. Robert Tisdale]

Shouldn't the compiler complain if you're not doing:

int main(int argc, char* argv[])

It probably should complain.
Why?

cat f.c

int f(char* s[]) {
char** p = s;
}

gcc -Wall -std=c99 -pedantic -c f.c

f.c: In function `f':
f.c:2: warning: unused variable `p'
f.c:3: warning: control reaches end of non-void function

Your compiler complains about you having declared an automatic variable
without using it and to not return a value from a function that is
declared as returning int. This has no relation to the given problem.

C performs an "implicit conversion"

from an array to a pointer to its first element

Right, when used as a value or formal parameter.

and from a pointer to an array.

Huh?
ITYM one can apply the '[]' operator to a pointer like
one would do with an array, so that:

int i = 21;
int a[42];
int *p = a;

leads to:

p == a;

I can write

cat f.c

int f(char* s[]) {
char** p = s;
return f(p);
}

gcc -Wall -std=c99 -pedantic -c f.c

and my compiler will not complain.

Yes, because this time you use p *and* you (formally) return a
value while in fact f() won't return anything at all as it will
crash your program when called because of infinite recursion.

Now that you've caught up with everyone else, please explain

> expand f.c

#include <stdio.h>

int f(char* s[10]) {
char** p = s;
fprintf(stdout, "%d = sizeof(s)\n", sizeof(s));
fprintf(stdout, "%d = sizeof(p)\n", sizeof(p));
return 0;
}

int main(int argc, char* argv[]) {
char * s[10];
fprintf(stdout, "%d = sizeof(s)\n", sizeof(s));
return f(s);
}

> gcc -Wall -std=c99 -pedantic -o main f.c
> ./main

40 = sizeof(s)
4 = sizeof(s)
4 = sizeof(p)

 

[Arthur J. O'Dwyer]

All I can see is that on your system
10*sizeof(char*) == 40 and sizeof(char**) == 4.
Not very surprisingly, I have to add.

Where's the joke?

Tisdale is a troll. Please just correct his posts quietly and
move away; or just ignore them, as I've learned to do. (And
yes, the "is he really at NASA?" question has been beaten to
death several times already. Check Google Groups.)

-Arthur

 

[Richard Heathfield]

Now that you've caught up with everyone else,

Don't be unpleasant, especially when you're wrong.

please explain
Delighted.

Syntax error.

#include <stdio.h>

int f(char* s[10]) {
char** p = s;
fprintf(stdout, "%d = sizeof(s)\n", sizeof(s));
fprintf(stdout, "%d = sizeof(p)\n", sizeof(p));

At the very least, on systems where a size_t is larger than an int the
behaviour is undefined.

Note that your demo program doesn't meet the requirements of the question,
which nowhere mentions char *s[10]. Still, never mind that. Observe:

rjh@tux:~/scratch> cat foo.c
#include <stdio.h>

void foo(char **r, char *s[], char *t[10])
{
printf("sizeof r = %lu\n", (unsigned long)sizeof r);
printf("sizeof s = %lu\n", (unsigned long)sizeof s);
printf("sizeof t = %lu\n", (unsigned long)sizeof t);
}

int main(int argc, char **argv)
{
if(argc >= 10)
{
foo(argv, argv, argv);
}
else
{
char *tmp[10];
foo(argv, argv, tmp);
}
return 0;
}

rjh@tux:~/scratch> ./foo 1 2 3 4 5 6 7 8 9 10
sizeof r = 4
sizeof s = 4
sizeof t = 4


Now for some history. Here are the first few lines of the source code from a
***very*** early C compiler. Regular readers will note the irony of my
quoting from it:

/* C compiler

Copyright 1972 Bell Telephone Laboratories, Inc.

*/

ossiz 250;
ospace() {} /* fake */

init(s, t)
char s[]; {
extern lookup, symbuf, namsiz;
char symbuf[], sp[];
int np[], i;

i = namsiz;
sp = symbuf;
while(i--)
if ((*sp++ = *s++)=='\0') --s;
np = lookup();
*np++ = 1;
*np = t;
}

Note the usage of np, which is defined as int np[], and used like this:
*np++ = 1;

Clearly, this is pointer syntax. [] was, in fact, used for pointer notation
early in C's development. Eventually, it was changed, but the usage of []
in function parameter lists lives on. It /still/ means pointer, in that
context.

Brian W Kernighan writes, in K&R2 starting at the foot of p99, "As formal
parameters in a function definition, char s[]; and char *s; are equivalent;
we prefer the latter because it says more explicitly that the parameter is
a pointer."

I don't expect Mr Tisdale to understand this, of course, but some other
people might find it interesting.

 

[E. Robert Tisdale]

Note that your demo program doesn't meet the requirements of the question,
which nowhere mentions char *s[10]. Still, never mind that. Observe:

> cat foo.c
#include <stdio.h>

void foo(char **r, char *s[], char *t[10]) {
printf("sizeof r = %lu\n", (unsigned long)sizeof r);
printf("sizeof s = %lu\n", (unsigned long)sizeof s);
printf("sizeof t = %lu\n", (unsigned long)sizeof t);
}

int main(int argc, char* argv[]) {
if(argc >= 10) {
foo(argv, argv, argv);
}
else {
char *tmp[10];
foo(argv, argv, tmp);
}
return 0;
}

> ./foo 1 2 3 4 5 6 7 8 9 10
sizeof r = 4
sizeof s = 4
sizeof t = 4

Now for some history.
Here are the first few lines of the source code
from a ***very*** early C compiler.
Regular readers will note the irony of my quoting from it:

/* C compiler

Copyright 1972 Bell Telephone Laboratories, Inc.

*/

ossiz 250;
ospace() {} /* fake */

init(s, t)
char s[]; {
extern lookup, symbuf, namsiz;
char symbuf[], sp[];
int np[], i;

i = namsiz;
sp = symbuf;
while(i--)
if ((*sp++ = *s++)=='\0') --s;
np = lookup();
*np++ = 1;
*np = t;
}

Note the usage of np, which is defined as int np[], and used like this:
*np++ = 1;

Clearly, this is pointer syntax. [] was, in fact, used for pointer notation
early in C's development. Eventually, it was changed
but the usage of [] in function parameter lists lives on.
It /still/ means pointer, in that context.

Brian W Kernighan writes, in K&R2 starting at the foot of p99,
"As formal parameters in a function definition,
char s[]; and char *s; are equivalent; we prefer the latter
because it says more explicitly that the parameter is a pointer."

That's pretty good.
Now, perhaps you are ready to answer Bret's question,
"Shouldn't the compiler complain if you're not doing:

int main(int argc, char* argv[])?"

The answer is that it should. I don't know why it doesn't.
It appears that you believe that
the reason has something to do with legacy code
which treats T t[] as pointer syntax like T t*
*in a function argument list*.
Please cite and quote the relevant *rational*
from the ANSI/ISO C standard if you can.

I don't expect Mr Tisdale to understand this, of course,
but some other people might find it interesting.

Don't be unpleasant, especially when you're wrong.

 

[E. Robert Tisdale]

Tisdale is a troll. Please just correct his posts quietly
and move away; or just ignore them, as I've learned to do.
(... Check Google Groups.)

I checked Google Groups and searched for

Dwyer troll

in the comp.lang.c newsgroup.
It seems that, whenever you lose and argument,
you accuse your opponent of being a troll. ;-)

 

[Irrwahn Grausewitz]

Please cite and quote the relevant *rational*
from the ANSI/ISO C standard if you can.

Read JTC1/SC22/WG14 N843 6.7.5.3#6 if you can,
then rip out the page, eat it & choke.

 

[Richard Heathfield]

E. Robert Tisdale wrote:

Now, perhaps you are ready to answer Bret's question,
"Shouldn't the compiler complain if you're not doing:

int main(int argc, char* argv[])?"

The compiler may issue a diagnostic for any reason it likes, but it is not
required to issue a diagnostic for int main(int argc, char **argv).

The answer is that it should.

It is not required so to do. See the Standard. The usual section.

I don't know why it doesn't.

I do.

It appears that you believe that
the reason has something to do with legacy code
which treats T t[] as pointer syntax like T t*
*in a function argument list*.

That is not (precisely) the reason that no diagnostic is required. The
reason no diagnostic is required is that no syntax error or constraint
violation is involved.

Please cite and quote the relevant *rational*
from the ANSI/ISO C standard if you can.

See 5.1.2.2.1 of ISO/IEC 9899:1999.

Don't be unpleasant, especially when you're wrong.

Nice try, but I'm not wrong, and I wasn't being unpleasant. I was simply
pointing out, in that first clause, that I don't think you are capable of
understanding what I had written. It appears, from your reply, that I was
pretty close to the mark.

 

[j]

Note that your demo program doesn't meet the requirements of the question,
which nowhere mentions char *s[10]. Still, never mind that. Observe:

> cat foo.c
#include <stdio.h>

void foo(char **r, char *s[], char *t[10]) {
printf("sizeof r = %lu\n", (unsigned long)sizeof r);
printf("sizeof s = %lu\n", (unsigned long)sizeof s);
printf("sizeof t = %lu\n", (unsigned long)sizeof t);
}

int main(int argc, char* argv[]) {
if(argc >= 10) {
foo(argv, argv, argv);
}
else {
char *tmp[10];
foo(argv, argv, tmp);
}
return 0;
}

> ./foo 1 2 3 4 5 6 7 8 9 10
sizeof r = 4
sizeof s = 4
sizeof t = 4

Now for some history.
Here are the first few lines of the source code
from a ***very*** early C compiler.
Regular readers will note the irony of my quoting from it:

/* C compiler

Copyright 1972 Bell Telephone Laboratories, Inc.

*/

ossiz 250;
ospace() {} /* fake */

init(s, t)
char s[]; {
extern lookup, symbuf, namsiz;
char symbuf[], sp[];
int np[], i;

i = namsiz;
sp = symbuf;
while(i--)
if ((*sp++ = *s++)=='\0') --s;
np = lookup();
*np++ = 1;
*np = t;
}

Note the usage of np, which is defined as int np[], and used like this:
*np++ = 1;

Clearly, this is pointer syntax. [] was, in fact, used for pointer notation
early in C's development. Eventually, it was changed
but the usage of [] in function parameter lists lives on.
It /still/ means pointer, in that context.

Brian W Kernighan writes, in K&R2 starting at the foot of p99,
"As formal parameters in a function definition,
char s[]; and char *s; are equivalent; we prefer the latter
because it says more explicitly that the parameter is a pointer."

That's pretty good.
Now, perhaps you are ready to answer Bret's question,
"Shouldn't the compiler complain if you're not doing:

int main(int argc, char* argv[])?"

You can either use ``char *argv[]'' or ``char **argv'' or if not
passing any arguments to main, then ``int main(void)''. In all cases
the compiler won't complain. Although this brings me to wondering
about something. In section 6.11.6 of c99 it says:

"6.11.6 Function declarators
1 The use of function declarators with empty parentheses (not
prototype-format parameter
type declarators) is an obsolescent feature."

So something like:

int foo(); would be considered obsolescent?


Also, would:

int main() /* This falls in 6.11.6? */
{
return 0;
}

The second example of main, it being part of a function definition, is
what is throwing me off and am unsure if it is considered obsolescent
as stated in 6.11.6.

The answer is that it should. I don't know why it doesn't.
It appears that you believe that
the reason has something to do with legacy code
which treats T t[] as pointer syntax like T t*
*in a function argument list*.
Please cite and quote the relevant *rational*
from the ANSI/ISO C standard if you can.

I don't expect Mr Tisdale to understand this, of course,
but some other people might find it interesting.

Don't be unpleasant, especially when you're wrong.

 

[E. Robert Tisdale]

You can either use ``char *argv[]'' or ``char **argv''

or if not passing any arguments to main, then ``int main(void)''.
In all cases the compiler won't complain.
Although this brings me to wondering about something.
In section 6.11.6 of c99 it says:

"6.11.6 Function declarators
1 The use of function declarators with empty parentheses
(not prototype-format parameter type declarators)
is an obsolescent feature."

So something like:

int foo();

would be considered obsolescent?
Also, would:

int main(void) { /* This falls in 6.11.6? */
return 0;
}

The second example of main,
it being part of a function definition, is what is throwing me off
and am unsure if it is considered obsolescent as stated in 6.11.6.

I don't know.
Function main is a special case.
The application isn't allowed to call it.
Your definition above would simply ignore any arguments passed to it.

No doubt, several of our indigenous "spiritualists"
will cite and quote the ANSI/ISO C "scripture"
as the reason why but I don't think any of them know
the *rationale* behind the dictates of the ANSI/ISO C standard.
Part of the problem is that the reasoning behind decision
taken collectively are seldom documented in great any detail
and are lost as time passes. The result is that
some of what appears in the standard is ritualistic
based upon superstitious belief.

 

[Kevin Easton]

You can either use ``char *argv[]'' or ``char **argv''

or if not passing any arguments to main, then ``int main(void)''.
In all cases the compiler won't complain.

The compiler doesn't have to complain about *any* incorrect declaration
of main - but the three you listed are all correct.

Yes.

No, that's not obsolescent, because it's using a prototype-format
parameter list. On the other hand, this is obsolescent:

int main()
{
return 0;
}

(It doesn't matter whether it's a definition or a declaration).

I don't know.
Function main is a special case.
The application isn't allowed to call it.

No, there's nothing wrong with the program calling its own main()
function.

Your definition above would simply ignore any arguments passed to it.

No doubt, several of our indigenous "spiritualists"
will cite and quote the ANSI/ISO C "scripture"
as the reason why but I don't think any of them know
the *rationale* behind the dictates of the ANSI/ISO C standard.
Part of the problem is that the reasoning behind decision
taken collectively are seldom documented in great any detail
and are lost as time passes. The result is that
some of what appears in the standard is ritualistic
based upon superstitious belief.

The reason behind obsoleting non-prototyped function declarations and
definitions is quite clear - because it removes the compiler's ability
to type-check function calls, and it gains nothing.

- Kevin.

 

[Richard Heathfield]

Right.

No, it doesn't. The keyword "void" fills the gap. In other words, this is a
function prototype - a declaration that explicitly documents the types of
its parameters, with void being used in this case to document that there
are /no/ parameters.

I don't know.

I do. int main(void) is not obsolescent.

Function main is a special case.

Only in that it is the entry point to the program and must therefore have a
carefully-defined interface.

The application isn't allowed to call it.

What are you wittering on about now? Of course the application is allowed to
call it.

 

[j]

No, it doesn't. The keyword "void" fills the gap. In other words, this is a
function prototype - a declaration that explicitly documents the types of
its parameters, with void being used in this case to document that there
are /no/ parameters.

Hm, in my original question I didn't have a ``void''.. not sure how that
ended up there.

 

[Irrwahn Grausewitz]

Hm, in my original question I didn't have a ``void''.. not sure how that
ended up there.

It's not magic. It's E.R.T.!

 

[Arthur J. O'Dwyer]

E. Robert Tisdale wrote: ^^^^^^^^^^^^^^^^^

j wrote:

In section 6.11.6 of c99 it says:

"6.11.6 Function declarators
1 The use of function declarators with empty parentheses
(not prototype-format parameter type declarators)
is an obsolescent feature."
int main(void) { /* This falls in 6.11.6? */

^^^^

No, it doesn't. [...]

Hm, in my original question I didn't have a ``void''.. not sure how that
ended up there.

This is *exactly* why people should reference the OP's question
directly, and ignore Tisdale's "contributions." In this case,
Tisdale modified your question to say something completely
different, and then two people separately quoted Tisdale's version
without looking at the real post again. (Don't feed the troll,
blah blah blah...)

BTW, Richard, I like that verb "wittering"!

-Arthur

 

[John Bode]

I'm curious why char** argv is acceptable in the main() declaration.

In the comp.lang.c FAQ (question 6.18) it says that pointers to
pointers and pointers to an array are not interchangable. However the
declaration:

int main(int argc, char** argv)

is common.

How does that work out? Shouldn't the compiler complain if you're not
doing:

int main(int argc, char* argv[])

Thanks,
Bret

First of all, the declarator char *argv[] declares argv to be an array
of pointers to char, not a pointer to an array of char.

char a[10]; /* a is a 10-element array of char */
char *a[10]; /* a is a 10-element array of pointer to char */
char (*a)[10]; /* a is a pointer to a 10-element array of char */
char *(*a)[10]; /* a is a pointer to a 10-element array of pointer to
char */

Secondly, in the context of a formal parameter declaration, char **x
and char *x[] are understood to mean the same thing. You cannot pass
a whole array to a function in C (this includes the main() function);
what actually happens is that a pointer to the first element of the
array is passed to the function. Therefore, char **x more correctly
describes what is happening. The [] notation is allowed to imply to
whomever is reading the program that the argument represents an array
of something, not just a pointer to an individual element. As far as
the code is concerned, however, all that's being passed is a pointer
to a single item (the first element of the array).

FWIW, this means you can't use sizeof to determine the array size in
the called function:

int main (void)
{
int x[10];

/* sizeof x == sizeof int * 10 */
foo (x); /* actually passes &x[0] */
return 0;
}

void foo (int *x)
{
/* sizeof x == sizeof int* */
}