5

I'm getting error like "expected an statement"

my code is as follows

#define IN_Tamper    0X00001000     /*P2.12 = EINT2*/
#define DIR_IN_Tamper    { FIO2DIR &= ~0X00001000 ; } 

/* main */
DIR_IN_Tamper(); 
if(((IN_Tamper >> 12) & 0x01) == 1)
            BUZZER_ON();
         else
            BUZZER_OFF();

I'm getting error saying

  1. Expected an statement for DIR_IN_Tamper();

  2. expected a statement for the else part.....

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6
  • or defined like: #define DIR_IN_Tamper(FIO2DIR) { FIO2DIR &= ~0X00001000 ; }; and call like DIR_IN_Tamper(FIO2DIR); what is FIO2DIR ? Commented Jul 12, 2013 at 10:29
  • #define DIR_IN_Tamper { FIO2DIR &= ~0X00001000 ; } like this...... Commented Jul 12, 2013 at 10:31
  • No I mean what is FIO2DIR ? if its macro then consider @phihag's answer. If its variable you wants to pass then defined like macro function as I suggested. Commented Jul 12, 2013 at 10:33
  • FIO2DIR is ARM7's key word... Commented Jul 12, 2013 at 10:38
  • 1
    lastly I have removed () from every where n its working fine.. thank you all.. Commented Jul 12, 2013 at 11:39

5 Answers 5

Reset to default
7

The C preprocessor is (at least in the way you use it) just a simple search-and-replace, so you're effectively running

/* main */
{ FIO2DIR &= ~0X00001000 ; } ();

This doesn't make any sense. Remove the parentheses in the line

DIR_IN_Tamper();

For BUZZER_ON and BUZZER_OFF, you want to remove the parentheses as well. If the macro isn't enclosed in curly braces, you also want to add those, like

if(((IN_Tamper >> 12) & 0x01) == 1) {
    BUZZER_ON
} else {
    BUZZER_OFF
}
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2 Comments

for BUZZER_ON/OFF I have used macro
like #define DIR_BUZ() FIO0DIR |= 0x10000000 ; //DIRECTION FOR BUZ PIn P0.28 #define B_BUZ_E(X) {(X==SET)? (FIO0SET |= PINO_BUZ) : (FIO0CLR |= PINO_BUZ) ;} #define BUZZER_ON() { DIR_BUZ(); B_BUZ_E(1); } #define BUZZER_OFF() { DIR_BUZ(); B_BUZ_E(0); }
2

DIR_IN_Tamper is defined as { FIO2DIR &= ~0X00001000 ; }, therefore when the preprocessor parses your code, this line

DIR_IN_Tamper();

is converted into

{ FIO2DIR &= ~0X00001000 ; }()

Which is clearly not correct. Not sure what exactly you're trying to achieve, but removing the parentheses will eliminate the syntax error:

DIR_IN_Tamper

Further to it, I suspect you have similar issues with BUZZER_ON and BUZZER_OFF.

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5 Comments

Shouldn't DIR_IN_Tamper rather be defined as a function-like macro?
Maybe it should but I see the question unclear in terms of what he exactly wants to do
@undur_gongor Maybe it should be, but, as KiaMorot pointed out, it's rather unclear from the question what the OP is trying to do. I showed one way to syntax errors.
if I remove {} it gives error error: #109: expression must have (pointer-to-) function type
@user2571585: You have to remove the normal parentheses (()), not the curly braces.
2

If you want to use DIR_IN_Tamper like a function, you need a function-like macro:

#define DIR_IN_Tamper()    { FIO2DIR &= ~0X00001000 ; }

Then, a better way to do it is:

#define DIR_IN_Tamper()    do { FIO2DIR &= ~0X00001000 ; } while(0)

... but that's a different story.

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7 Comments

Yes, it is a macro. But you try to "invoke" it like a function (with ()). Therefore you need a function-like macro.
macro of BUZZER_OFF is as follows-
#define BUZZER_OFF() { DIR_BUZ(); B_BUZ_E(0); }
do I need to remove () or {} here also..?? its not single statement but.
First and foremost, you should add the definitions of the other macros to your question.
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1

Single-statement, function-like macros

Please do not use curly braces ({ and }) when defining single-statement macros like DIR_IN_Tamper.

To safely define a function-like macro, simply put your definition between parentheses, like this:

#define DIR_IN_Tamper() (FIO2DIR &= ~0X00001000)

Then, call your macro like this:

DIR_IN_Tamper();

It will behave like a functions which changes the value of FIO2DIR and then returns the changed value:

/* Your macro rewritten as a function.
   The return type should be the type of FIO2DIR */
uint32_t DIR_IN_Tamper()
{
    return (FIO2DIR &= ~0X00001000);
}

Multi-statement, function-like macros

If you ever need to define a multi-statement macro, see this other C FAQ entry.

For example, define BUZZER_OFF as:

#define BUZZER_OFF() do { DIR_BUZ(); B_BUZ_E(0); } while (0)
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5 Comments

Armaccess.c(973): error: #109: expression must have (pointer-to-) function type
@user2571585: You have to remove the normal parentheses (()), not the curly braces.
@user2571585 I commented something about {} to your question
I have removed () braces from main.. now it looks like BUZZER_OFF; but still error in else
When in doubt, look at the pre-processed source: see stackoverflow.com/q/3742822/238421.
0

Macros in C are not functions. DIR_IN_Tamper(); should be DIR_IN_Tamper;.

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6 Comments

It is like this #define DIR_IN_Tamper FIO2DIR &= ~0X00001000 & not like u have mentioned..
Huh? I mean the call. Not the macro definition. Seems you're too much confused about C. You should get a decent C book and read it before writing code.
Then make a question about your new error, or say what error it is.
code says as follows DIR_IN_Tamper; if(((IN_Tamper >> 12) & 0x01) == 1) BUZZER_ON; else BUZZER_OFF; where in BUZZER_OFF is again a macro defined as #define BUZZER_OFF { DIR_BUZ; B_BUZ_E(0); } since its group of statements I have used { } braces...
Mu advice is that you avoid macros in C, they're more complicated and prone to errors than you might think. This explains why.
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