Re: const char ** syntax question
On Oct 12, 2:23 am, sal...@icdc.com wrote:
On Oct 11, 2:01 am, "Jim Langston" <tazmas...@rocketmail.com> wrote:
<sal...@icdc.com> wrote in message
news:1192030356.354157.20440@22g2000hsm.googlegroups.com...
What is wrong with line 4 of the program shown below?
Visual Studio 7.1 compiles it without incident. gcc 4.1.0 reports an
error.
int main()
{
char * p = "abc";
const char ** paddr = &p;
return 0;
}
It's the old "Just what is constant?" question. Change it to
char* const * paddr = &p;
and it will compile.
Our software includes several open source libraries,
and runs on at least six platforms. An interface to
one open source library takes
(in an early version) char ** p
(in a later version) const char ** p
I would like a way to write code that operates with
either version. (Running the same software on all
machines is as difficult as synchronizing clocks.)
You could write a wrapper class that would match either parameter
declaration - if you really wanted to go to such lengths:
struct CPtrPtr
{
CPtrPtr(char **p) : p_(p) {}
operator char **() { return p_; }
operator const char **()
{
return (const char **) p_;
}
private:
char **p_;
};
and sample usage:
void f1(char **p) {}
void f2(const char **p) {}
int main()
{
char *p;
CPtrPtr paddr = &p;
f1(paddr);
f2(paddr);
// or
f1( CPtrPtr(&p));
f2( CPtrPtr(&p));
}
Greg
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