// man inet_net_ntop
/* Link with "-lresolv" */
 
#include <arpa/inet.h>
#include <stdio.h>
#include <stdlib.h>
 
#define errExit(msg)    do { perror(msg); exit(EXIT_FAILURE); \
                        } while (0)
 
int
main(int argc, char *argv[])
{  
 

    The whole idea of using 'do/while' version is to make a macro which will expand into a regular statement, not into a compound statement. This is done in order to make the use of function-style macros uniform with the use of ordinary functions in all contexts.

    Consider the following code sketch:

    if (<condition>)
      foo(a);
    else
      bar(a);

    where foo and bar are ordinary functions. Now imagine that you'd like to replace function foo with a macro of the above nature [named CALL_FUNCS]:

    if (<condition>)
      CALL_FUNCS(a);
    else
      bar(a);

    Now, if your macro is defined in accordance with the second approach (just { and }) the code will no longer compile, because the 'true' branch of if is now represented by a compound statement. And when you put a ; after this compound statement, you finished the whole if statement, thus orphaning the else branch (hence the compilation error).

    One way to correct this problem is to remember not to put ; after macro "invocations":

    if (<condition>)
      CALL_FUNCS(a)
    else
      bar(a);

    This will compile and work as expected, but this is not uniform. The more elegant solution is to make sure that macro expand into a regular statement, not into a compound one. One way to achieve that is to define the macro as follows:

    #define CALL_FUNCS(x) \
    do { \
      func1(x); \
      func2(x); \
      func3(x); \
    } while (0)

    Now this code:

    if (<condition>)
      CALL_FUNCS(a);
    else
      bar(a);

    will compile without any problems.

    However, note the small but important difference between my definition of CALL_FUNCS and the first version in your message. I didn't put a ; after } while (0). Putting a ; at the end of that definition would immediately defeat the entire point of using 'do/while' and make that macro pretty much equivalent to the compound-statement version.

    I don't know why the author of the code you quoted in your original message put this ; after while (0). In this form both variants are equivalent. The whole idea behind using 'do/while' version is not to include this final ; into the macro (for the reasons that I explained above).