Java-style println() in C

**TheBlackAdder** · 05-20-2017

Hi,

I've been trying to create a java-style System.out.println() function in C, so that I don't have to keep writing \n at the end of every printf(), which I also keep forgetting...

I've read this thread about it:
Writing Println code
Where you concluded it not to be possible in C.
It is.

You just have to use a macro to do it:

Code:

#define println(fmt, args...) ({\
    int __a = printf(fmt,args);\
    printf("\n");\
    ++__a;\
})

It's consistent with printf() in the way that the return value is the number of total characters printed.

Hope it's helpful to others who want to do this.

**Salem** · 05-20-2017

Except it doesn't work.

Code:

#include <stdio.h>

#define println(fmt, args...) ({\
    int __a = printf(fmt,args);\
    printf("\n");\
    ++__a;\
})

int main ( ) {
  printf("hello\n");
  println("hello");
  return 0;
}

$ gcc foo.c
foo.c: In function ‘main’:
foo.c:4:30: error: expected expression before ‘)’ token
     int __a = printf(fmt,args);\
                              ^
foo.c:11:3: note: in expansion of macro ‘println’
   println("hello");
   ^
$ gcc -std=c99 foo.c
foo.c: In function ‘main’:
foo.c:4:30: error: expected expression before ‘)’ token
     int __a = printf(fmt,args);\
                              ^
foo.c:11:3: note: in expansion of macro ‘println’
   println("hello");
   ^
$ gcc -std=c11 foo.c
foo.c: In function ‘main’:
foo.c:4:30: error: expected expression before ‘)’ token
     int __a = printf(fmt,args);\
                              ^
foo.c:11:3: note: in expansion of macro ‘println’
   println("hello");
   ^

**whiteflags** · 05-21-2017

I suppose you could create some nonsense like this - but the ratio of effort spent to effect is not that great.

Code:

#include <stdio.h>
#include <stdarg.h>

int jck_println ( const char * format, ... )
{
    int result = 0;
    char buffer[BUFSIZ];
    va_list args;
    
    va_start(args, format);
    result = vsprintf(buffer, format, args);
    va_end(args);
    
    if (result > 0) {
        result = puts ( buffer );
    }

    return result;
}

int main(void)
{
    jck_println("%d", 123);
    jck_println("hello");
    jck_println("%s %d", "hello", 2);

    return 0;
}

It's still not all that java-like at all. Not to mention that it is assumed in this code that no one will ever try to print more than about BUFSIZ characters at once. It is probably safer and easier just to use "\n" in your format strings.

**c99tutorial** · 05-24-2017

Here is an alternative. All you need to do is call vprintf followed by some code which outputs a newline. You don't need buffers or anything like that. If you are using GCC you can even get it to check your format string for you.

Code:

#include <stdio.h>
#include <stdarg.h>

#ifdef __GNUC__
int printfln(const char *format, ...)
         __attribute__ ((format (printf, 1, 2)));
#endif

int printfln(const char *format, ...)
{
    va_list ap;
    va_start(ap, format);
    int ret = vprintf(format, ap);
    va_end(ap);
    puts("");
    return ret;
}

//
// Example usage
//

int main()
{
    printfln("x: %d"); // WARNING: 'int' argument missing.
    printfln("y: %d", 234);
    printfln("z: %d", 345);
    printfln("Bye.");
    return 0;
}

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