how to convert ads to iar?

I'm sure lots of consultants could, for the usual hourly consideration. How much are you paying?

best regards Pete

In article , lzh08 writes

Lots of people. I could. Why should I?

Try

char one = 'a'; char two ='d'; char three ='s'; char four = 8; char five = 3; char six =1;

one = 0ne + four; two = two -five; three = three - six;

ADS is RF design suit. IAR is embedded workbench. It wouldn't be too simple to make a conversion...

Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

In article , lzh08 writes

Run this.... not pretty but it converts ads to iar ! (I even used an IAR EWB to do it :-)

#include

int main(void ) {

char one = 'a'; char two ='d'; char three ='s';

char four = 8; char five = 3; char six =1;

printf("%c %c %c \n", one, two, three);

one = one + four; two = two -five; three = three - six;

printf("%c %c %c \n", one, two, three);

return 0; }

Nope. ARM Developer Suite. ;)

pete

In article , Chris Hills writes

/* how to convert ads to iar? */ /* the invoice is in the post */ /* next converting ghs to iar */ /* "keil" to "iar" will need some thinking about */ /* Some one else can do "windriver" to "iar" :-) */

#include

int main(void ) {

char one = 'a'; char two ='d'; char three ='s';

char four = 8; char five = 3; char six =1;

printf("%c %c %c \n", one, two, three);

one = one + four; two = two -five; three = three - six;

printf("%c %c %c \n", one, two, three);

return 0; }

I think he knows that. He is just being perverse like the rest of us.

Anyway I provided a hard coded solution in C using EWB :-)

No problem.

#include int main() { char x[6]; fgets(x, 6, stdin); printf("%c%c%c\n", x[2], x[1] - 4, x[0]+x[3]-x[1]); return 0; }

Even works for upper-case :-)

Stefan

Any one want to try "windriver" to" iar"? :-)

Without knowing what you want to convert then no-one can help you. If you've written ANSI C then everything will run just dandy. However, if you've used compiler extensions then you need to find the appropriate extensions under IAR. Their support people are very helpful (at a price) and respond quickly.

If you write code that conforms to standards and liberally document the places where it doesn't then you won't get this problem in the future.

#include int main() { char x[11]; char *p = &x[1]; fgets(x, 11, stdin); printf("%c", *p); p += 2; printf("%c", *p - 3); p += 5; printf("%c\n", *p); return 0; }

... snip ...

Fails if the char. set is ebcdic (or some others) :-)