Syntax error on #define

[Anshul]

Hi,

I tried compiling this code -

int main()
{
char abc[10] = "%s";
printf("\033[1m" abc "\033[m" , "doing");
}

with gcc version 2.96. I get this error message -

parse error before `abc'

Please explain, whats wrong here. The intention is to print a string
in bold characters on the screen.

Thanks,
Anshul

 

[Anshul]

Oops, bad subject line.

Thanks,
Anshul

 

[santosh]

Hi,

I tried compiling this code -

int main()
{
char abc[10] = "%s";
printf("\033[1m" abc "\033[m" , "doing");

Only adjacent string literals are concactenated. The expression abc is
not a string literal, but a pointer constant value.

Try:

printf("\033[1m", " %s ", "\033[m\n", "doing");

 

[Richard Heathfield]

santosh said:

Hi,

I tried compiling this code -

int main()
{
char abc[10] = "%s";
printf("\033[1m" abc "\033[m" , "doing");

Only adjacent string literals are concactenated. The expression abc is
not a string literal, but a pointer constant value.

Try:

printf("\033[1m", " %s ", "\033[m\n", "doing");

Er, no.

printf("\033[1m%s\033[m\n", "doing");

or even just:

printf("\033[1mdoing\033[m\n");

(although the former version is perhaps slightly more maintainable, in
that it's easier to see what needs changing when the need arises).

 

[Joachim Schmitz]

Hi,

I tried compiling this code -

int main()
{
char abc[10] = "%s";
printf("\033[1m" abc "\033[m" , "doing");

printf("\033[1m%s\033[m" , "doing");
fflush(stdout);
return 0;

}

with gcc version 2.96. I get this error message -

parse error before `abc'

Please explain, whats wrong here. The intention is to print a string
in bold characters on the screen.

Thanks,
Anshul

Bye, Jojo

 

[Joachim Schmitz]

Hi,

I tried compiling this code -

int main()
{
char abc[10] = "%s";
printf("\033[1m" abc "\033[m" , "doing");

Only adjacent string literals are concactenated. The expression abc is
not a string literal, but a pointer constant value.

Try:

printf("\033[1m", " %s ", "\033[m\n", "doing");

Still doesn't work, got to be:
printf("\033[1m%s\033[m\n", "doing");

Bye, Jojo

 

[Richard Tobin]

Only adjacent string literals are concactenated. The expression abc is
not a string literal, but a pointer constant value.

Try:

printf("\033[1m", " %s ", "\033[m\n", "doing");

^ ^

Take out those commas!

-- Richard

 

[santosh]

Hi,

I tried compiling this code -

int main()
{
char abc[10] = "%s";
printf("\033[1m" abc "\033[m" , "doing");

Only adjacent string literals are concactenated. The expression abc is
not a string literal, but a pointer constant value.

Try:

printf("\033[1m", " %s ", "\033[m\n", "doing");

Sorry to all about this. It should be:

printf("\033[1m" "%s" "\033[m\n", "doing");

or better yet, versions that Richard, Richard and Joachim posted.

 

[Dik T. Winter]

> >> Hi,
> >>
> >> I tried compiling this code -
> >>
> >> int main()
> >> {
> >> char abc[10] = "%s";
> >> printf("\033[1m" abc "\033[m" , "doing");

> >
> > Only adjacent string literals are concactenated. The expression abc is
> > not a string literal, but a pointer constant value.
> >
> > Try:
> >
> > printf("\033[1m", " %s ", "\033[m\n", "doing");

Eh, no. But the following *will* work:
#define abc "%s"
printf("\033[1m" abc "\033[m", "doing");