Why does semi-colon after a macro cause illegal else compile error

Question

Apparently I have a fundamental misunderstanding of how macros work. I thought a macro just caused the preprocessor to replace the @defined macro with the replacement text. But apparently that's not always the case. My code follows: TstBasInc.h

#pragma once

#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <stdint.h>
#include <stdbool.h>
#include <stdarg.h>
#include <time.h>

//  copy macro
#define Cpy(ToVar, FrmVar) do {                                                                                 \
                                errno = strncpy_s(ToVar, sizeof(ToVar), FrmVar, _TRUNCATE);                     \
                                    if (errno == STRUNCATE)                                                     \
                                        fprintf(stderr, "string '%s' was truncated to '%s'\n", FrmVar, ToVar);  \
                            } while(0);

//  clear numeric array macro
#define ClrNumArr(ArrNam, ArrCnt)           \
            for (s = 0; s < ArrCnt; s++)    \
                ArrNam[s] = 0;


uint32_t    s;          //  subscript

typedef struct {
    short   C;
    short   YY;
    short   MM;
    short   DD;
} SysDat;

TstMacCmpErr:

#include "stdafx.h"
#include "TstBasInc.h"      //  test basic include file

#define ARRCNT 3

int main()
{
    char    Cnd = 'E';      //  define to use 'else' path
    char    ToVar[7 + 1];   //  Cpy To-Variable
    int     IntArr[ARRCNT]; //  integer array

    Cpy(ToVar, "short")                     //  compiles with or without the semi-colon
    if (Cnd != 'E')
//      Cpy(ToVar, "short string");         //  won't compile:  illegal else without matching if
        Cpy(ToVar, "short string")          //  will compile
    else
        Cpy(ToVar, "extra long string");    //  compiles with or without the semi-colon
//  the following code shows how I thought the macro would expand to    
    {                                                                                   \
        errno = strncpy_s(ToVar, sizeof(ToVar), "short str", _TRUNCATE);                \
        if (errno == STRUNCATE)                                                         \
            fprintf(stderr, "string '%s' was truncated to '%s'\n", "short str", ToVar);;    \
    }

    if (Cnd == 'E') {
        ClrNumArr(IntArr, ARRCNT)               //  compiles with or without the semi-colon
            printf("intarr[0] = %d\n", IntArr[0]);
    }
    else
        printf("intarr[0] is garbage\n");

    return 0;
}

The results follow:

string 'extra long string' was truncated to 'extra l'
string 'short str' was truncated to 'short s'
intarr[0] = 0;

As the comments say, when I have a semi-colon after Cpy(ToVar, "short string"); it won't even compile because I get an "C2181 illegal else without matching if" error. As you see, I tried adding a do-while in the macro as suggested in this post, but it didn't make any difference. When the macro code is copied directly (even without the do-while) that code works okay. I would have thought just adding the braces in the macro would have fixed the problem but it didn't. It must have something to do with the Cpy ending in an if because the ClrNumArr macro compiles with or without the semi-colon. So can someone tell me why the Cpy macro doesn't just replace the text? I must be missing something simple.

I'm using VS 2015 Community Edition Update 1.

Edit: I documented the problem and isolated it to (I think) the if statement in the Cpy macro. Still nobody has explained why the macro didn't expand the way I thought it should. That should have been the title of the post, as that's my question more than the problem with the semi-colon, which I now have a solution for.

Answer 1

In case you put a semicolon,

    Cpy(ToVar, "short")                     //  compiles with or without the semi-colon
    if (Cnd != 'E')
//      Cpy(ToVar, "short string");         //  won't compile:  illegal else without matching if
        Cpy(ToVar, "short string")          //  will compile
    else
        Cpy(ToVar, "extra long string");    //  compiles with or without the semi-colon

this makes the syntax look like

if (...)
    ...;
    ;                // <-- The problem here
else
   ...;

which is illegal syntax, as the else will have no associated if in this syntax.

OTOH, in case you write

if (...) {
    ...;
    ;                // <-- no problem here, inside a block
 }
else
   ...;

the else is associated with the previous if. So, all fine.

Answer 2

The problem becomes obvious if you look at what happens at expansion. The macro already expans to a statement including trailing semicolon.

The problem becomes apparent if you're using it together with an if-else construct since if you append an extra semicolon it would expand into two statements.

For example if you do the following:

#define HELLO printf("Hello\n");

and then

if( some_condition )
    HELLO;
else
    something_else();

this would expand to

if( some_condition )
    printf("Hello\n");;
else
    something_else();

and you see that there's two statements between the if and else which means that the else becomes out of place.

The very reason of using the do - while(0) is to create something that would fit where one statement is to be fitted and then it's imperative that you leave out the trailing semicolon after the while(0). That is your define should look like this instead:

#define Cpy(ToVar, FrmVar) do   {                                                                                 \
                            errno = strncpy_s(ToVar, sizeof(ToVar), FrmVar, _TRUNCATE);                     \
                                if (errno == STRUNCATE)                                                     \
                                    fprintf(stderr, "string '%s' was truncated to '%s'\n", FrmVar, ToVar);  \
                        } while(0)

Answer 3

In a comment to Sourav's answer, you asked why this works as a macro body:

{ 
    errno = strncpy_s(ToVar, sizeof(ToVar), "short str", _TRUNCATE); 
    if (errno == STRUNCATE) 
        fprintf(stderr, "string '%s' was truncated to '%s'\n", "short str", ToVar);; 
}

That's because in your context, it evaluates to this:

if (Cnd != 'E')
{ 
    errno = strncpy_s(ToVar, sizeof(ToVar), "short str", _TRUNCATE); 
    if (errno == STRUNCATE) 
        fprintf(stderr, "string '%s' was truncated to '%s'\n", "short str", ToVar);; 
}
else
    Cpy(ToVar, "extra long string");    //  compiles with or without the semi-colon

The if part of the block, which previously was a single statement is now a block. So now the else is properly matched.

That's still not a good way to implement a macro, however. Then including the semicolon (which users of a macro would generally expect to work) would cause an error.

if (Cnd != 'E')
    Cpy(ToVar, "short string");     // invalid syntax
else
    Cpy(ToVar, "extra long string");

You keep saying "why didn't the macro expand the way I thought it would". It is expanding as you thought it would. Based on your original definition, this:

if (Cnd != 'E')
    Cpy(ToVar, "short string");
else
    Cpy(ToVar, "extra long string");    //  compiles with or without the semi-colon

Expands to this:

if (Cnd != 'E')
    do{
        errno = strncpy_s(ToVar, sizeof(ToVar), "short string", _TRUNCATE);
        if (errno == STRUNCATE)
            fprintf(stderr, "string '%s' was truncated to '%s'\n", "short string", ToVar);
    } while(0);;
else
    Cpy(ToVar, "extra long string");    //  compiles with or without the semi-colon

The first semicolon after the while(0) (which is part of the macro) ends the first part of the if statement. The second semicolon (not part of the macro) is another (empty) statement. This statement is considered to be outside of the if statement. Because of this, the else is misplaced.

You would get the same error if you had something like this:

if (Cnd != 'E')
    do{
        errno = strncpy_s(ToVar, sizeof(ToVar), "short string", _TRUNCATE);
        if (errno == STRUNCATE)
            fprintf(stderr, "string '%s' was truncated to '%s'\n", "short string", ToVar);
    } while(0);
fprintf(stderr, "this is outside of the if statement\n");
else
    Cpy(ToVar, "extra long string");    //  compiles with or without the semi-colon

Note that I indented the fprintf call to where is logically belongs. That's exactly where the empty statement (i.e. the extra semicolon) belongs as well.