Is there any setfill() alternative for C?

问题

In C++:

int main()
{
    cout << setfill('#') << setw(10) << 5 << endl;

    return 0;
}

Outputs:

#########5

Is there any setfill() alternative for C? Or how to do this in C without manually creating the string?

回答1:

No, there is no direct alternative.

But if you have a well-behaved snprintf (one that behaves as described by the C99 Standard), this works without creating a new string; creating only 2 temporary ints

#include <stdio.h>

int main(void) {
  int filler = '#'; /* setfill('#') */
  int width = 10;   /* setw(10)     */
  int target = 5;   /* 5            */

  /* ******** */
  int s = snprintf(NULL, 0, "%d", target);
  for (int i = 0; i < width - s; i++) {
    putchar(filler);
  }
  printf("%d\n", target);
  /* ******** */

  return 0;
}

EDIT: running version at ideone.

EDIT2: Differences between the C99 Standard's snprintf and Windows _snprintf (thank you for the link, Ben):

prototype: int snprintf(char *restrict buffer, size_t n, const char *restrict format, ...);
prototype: int _snprintf(char *buffer, size_t n, const char *format, ...);
snprintf writes no more than (n-1) bytes and a NUL
_snprintf writes no more than (n) bytes, the last of which may be NUL or other character
snprintf returns the number of characters needed for format (can be larger than n) or -1 in case of encoding error
_snprintf returns a negative value if n is not large for the string; or n if a NUL byte hasn't been written to buffer.

You can run the mis-behaving _snprintf in a loop, increasing n until you find the right value

/* absolutely not tested, written directly on SO text editor */
int i;
size_t rightvalue = 0;
char buffer[SOME_DEFAULT_VALUE];
do {
    if (sizeof buffer < rightvalue) /* OOPS, BIG BIG OOPS */;
    i = _snprintf(buffer, rightvalue, "%d", 42);
} while (i != rightvalue++);
/* rightvalue already has space for the terminating NUL */

回答2:

 int x= 5; 
 printf("%010d",x);

will output : 0000000005

Now if you really want '#' instead of '0' you'll have to replace them manually in the string.

Maybe :

char buf[11], *sp = buf; 
snprintf(buf, 11, "%10d", x); 
while( (sp = strchr(sp, ' ')) != '\0'){ *sp = '#'; }
puts(buf);

回答3:

Everybody wants to call printf twice... printf is one of the most expensive functions around.

Here:

char buffer[18] = "##########";
puts(buffer + snprintf(buffer+strlen(buffer), 8, "%d", 5));

回答4:

Not built in as standard

I would probably sprintf() the number to a string and get the count of characters then output the correct number of '#' first before printing the string.

回答5:

The following will do it using memset in C assuming a 1-byte char. It still creates a 'string', though I'm not sure how manually you don't want it to be.

int main(void)
{

   char buf[MAX_LENGTH];

   memset(buf, '#', 10);
   buf[10]='\0';
   printf("%s5\n", buf);
}

Depending what you want to actually do with it, you could dynamically create the buffer to be the appropriate size and return it from a helper function if you so desire.

来源：https://stackoverflow.com/questions/3867633/is-there-any-setfill-alternative-for-c

标签

c++

output-formatting