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I call my program like this:

$ ./test http://myserver.com/get?id=XaXz-dryq...wpxDB64juPIhQR

But only this get printed out:
./test
http://myserver.com/get?id=XaXz-dryqLg

It seems like the second 'argv' get chopped off at '&' , can you
please tell me why is that and how to fix it?

int main( int i_argc, char *argv[] )
{

fprintf (stderr, "******************\n");
for (int i = 0; i < i_argc; i++) {
fprintf (stderr, " %s\n",argv[i]);
}

//....
}

wrote:
> I call my program like this:
>
> $ ./test http://myserver.com/get?id=XaXz-dryq...wpxDB64juPIhQR
>
> But only this get printed out:
> ./test
> http://myserver.com/get?id=XaXz-dryqLg
>
> It seems like the second 'argv' get chopped off at '&' , can you
> please tell me why is that and how to fix it?
>
The problem is with your shell (it is interpreting the '&'), not your
program. Try quoting the input string.

--
Ian Collins.