error in code

Post your code

Post your code

#include <stdio.h>

int main( integer argc, char ***argv[] ) interrupt 10
{ ilv( 12 , 7); // chek the list

initibit( 11 ); // ???

// while (argc<>0) compiler thinks is bad

while (argc); { initibit( argv[ argc ] };

argc := argc - 1; }

printf( "program complete %d\a" , argc);

return 0

}

but i do not interupt
then what is this
int main( integer argc, char ***argv[] ) interrupt 10

Erik

i say already again

i do not clik the stop build button!!!

the code txes to pc. i think interrupt is for sbuf

sbuf sends to pc not comopiler

i thinking is big bug in compiler !!!!????

"when i build it gives a message

** error C138 interupt can not recieive or return a value"

The first thing to do when you get such a message is to read its description in the Manual:

"Description: An interrupt function was defined with one or more formal parameters or with a return value. Interrupt functions may not contain invocation parameters or return values."
http://www.keil.com/support/man/docs/c51/c51_c138.htm

So, let's have a look at your code:

int main( integer argc, char ***argv[] ) interrupt 10

The interrupt keyword defines the funcion to be an interrupt function:
http://www.keil.com/support/man/docs/c51/c51_le_interruptfuncs.htm

Clearly your code gives the function both a return value and some parameters, and you can't do that - just as the message says!

"the proj is needed and i now worry big in my pants"

Yes, I'm afraid you do have a very big worry here indeed!
You clearly have a lot of basic study to do about embedded programming in general and the Keil tools in particular...

"i say already again
i do not clik the stop build button!!!"

It's not talking about you interrupting the build process!

The concept of Imterrupts is absolutely basic to just about all computers - they go by the name of "IRQ" on PCs (short for "interrupt request")

www.8052.com/tutflow.phtml

If you don't understand this now, you have a very long way to go to complete your project!

:=0

"i thinking is big bug in compiler !!!!????"

No - just a big hole in your understanding

initibit( 11 );

Here, initibit() is being passed an integer.

initibit( argv[ argc ]

Despite the syntax errors, here initibit() is being a pointer to a pointer to a char pointer.

What is initibit() actually doing with the pointer -- erm, I mean integer?

argc := argc - 1;

Did you enable the Pascal compile option?