Re: Understanding char **argv

kevin.eugene08@xxxxxxxxxxxxxx wrote:
Hello all,

Forgive the somewhat simple question I am sure this will be, but I am
having some problem understanding what: char **argv looks like. For
instance, i know through trial and error that if I do this:

#include <stdio.h>

int main(int argc, char *argv[])
{
int l;
for( l=0; l<argc; l++ )
{
printf("Pointer: %p\tValue: %s\n", argv++, *argv);
}

return 0;
}

And i invoke the above as:

./foo one two three

Then I will see:


0x10202 foo
0x10204 one
0x10208 two
ox1020c three

Listed, but I would have expected to have needed to write:

*(*(argv))

So as to dereference what the pointer that *argv pointed to. Or have
I missed the point?

Maybe two points.

1. The parens are not needed. 'char **argv' is what it is, a pointer to a pointer to char.

2. In your printf statement above you have both argv++ and *argv. Because we have no way to know which expression might be evaluated first, this is classic Undefined Behavior.

#include <stdio.h>

int main(int argc, char **argv)
{
int l;
for (l = 0; l < argc; l++) {
printf("Pointer: %p\tValue: %s\n", argv, *argv);
argv++;
}
return 0;
}

...is the way I would do it.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
.

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