How to convert from a string to an expression?

[David]

for example:
int x=10;
char *str="\"x=%d\", x";

if I use
printf(expressi on(str));
then the results is
x=10

even I know it's not easy to return a expression from a function.

can u give me a good idea?

[Michael Mair]

David wrote:

for example:
int x=10;
char *str="\"x=%d\", x";

if I use
printf(expressi on(str));
then the results is
x=10

even I know it's not easy to return a expression from a function.

can u give me a good idea?

If you tell us what you want to achieve and give us the code
you have up to now or at least your representation of an "expression "
(which is no builtin C datatype).

Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Keith Thompson]

Michael Mair <Mi**********@i nvalid.invalid> writes:

David wrote:

for example:
int x=10;
char *str="\"x=%d\", x";

if I use
printf(expressi on(str));
then the results is
x=10

even I know it's not easy to return a expression from a function.

can u give me a good idea?

If you tell us what you want to achieve and give us the code
you have up to now or at least your representation of an "expression "
(which is no builtin C datatype).

I think he's done that. His representation of an expression seems to
be a string containing a sequence of arguments to printf, separated by
commas. He wants a function

char *expression (char *expr);

that will return the string that would have been printed by printf().

David, is that about right?

Unfortunately, what you're trying to do is pretty much impossible.
Given a string "x", there's no way in C to get the value of the
variable whose name happens to be x. Variable names appear in source
code, but typically do not appear in the executable program; even if
they do, there's no way to get at them. Roughly speaking, the
compiler decides where x is going to be stored; all references to x in
your program are transformed into references to that memory location,
and the name vanishes.

Now if you explicitly associate the string "x" with the variable,
something like
record_name("x" , &x);
you might be able to do something like what what you describe, but
it's going to be complicated. The above would not be adequate; you'll
also need some way to associate "x" with the type of x, which happens
to be int -- and there's no simple way to represent a type at runtime.

The first thing you should do is decide whether whatever problem
you're trying to solve can be solved in some other way.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Michael Mair]

Keith Thompson wrote:

Michael Mair <Mi**********@i nvalid.invalid> writes:

David wrote:

for example:
int x=10;
char *str="\"x=%d\", x";

if I use
printf(expressi on(str));
then the results is
x=10

even I know it's not easy to return a expression from a function.

can u give me a good idea?
If you tell us what you want to achieve and give us the code
you have up to now or at least your representation of an "expression "
(which is no builtin C datatype).

I think he's done that. His representation of an expression seems to
be a string containing a sequence of arguments to printf, separated by
commas. He wants a function

char *expression (char *expr);

that will return the string that would have been printed by printf().

Ah, I completely misread that. He sort of stringizes the arguments
he would normally pass to printf() and wants convert() to do
printf()s job but only with the string as input.

Hmmm...
[snip: Explanation of problems] Now if you explicitly associate the string "x" with the variable,
something like
record_name("x" , &x);
you might be able to do something like what what you describe, but
it's going to be complicated. The above would not be adequate; you'll
also need some way to associate "x" with the type of x, which happens
to be int -- and there's no simple way to represent a type at runtime.

Yep. Specialised solution:
One decides that a to z are "enough" and that all variables are
of either the largest integer or the largest floating point type.
Then you can store the addresses in an array:
void *vars['z'+1];

.....
unsigned long var[26], *ptmp, sink;
int i;
for (i=0, ptmp = var; i<='z'; i++)
if (isalpha(i) && islower(i))
vars[i] = ptmp++;
else
vars[i] = &sink;
(untested)
Alternatively, one could use NULL instead of &sink.
However, this is no general and no easily extendible _and_
maintainable solution.

Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.