sprintf and unsigned int array in C

Question

I have a pointer to an array of ints and the length of the array as such:

unsigned int length = 3;
int *array;        // Assume the array has 3 initialized elements

I also have a string and a buffer (assume it is sufficiently large) to put into sprintf as such:

char buffer[128];
const char *pattern = "(%d, %d, %d)\n";

Assume that pattern will only have "%d"s and other characters in it, but could be any form (i.e. "Test %d: %d" or "%d %d"), and that the length of array will always be the same as the number of "%d"s.

Since the length of the array can be anything, is there any way that I can do sprintf (buffer, pattern, &array[0], &array[1], &array[2]) without explicitly enumerating the elements of array? Something along the lines of sprintf (buffer, pattern, array). I can write as many helper functions as are necessary. I was thinking of faking a va_list, but this seems to be bad practice as it restricts the program to a certain compiler.

Sergey Kalinichenko · Accepted Answer · 2012-12-13 01:58:26Z

9

Passing all elements in a single va_list is not going to help, because the format string needs to be created in a loop anyway. Since you cannot escape the loop anyway, you might as well do the printing in the same loop:

int data[] = {12, 345, 6789, 101112};
char buf[128], *pos = buf;
for (int i = 0 ; i != 4 ; i++) {
    if (i) {
        pos += sprintf(pos, ", ");
    }
    pos += sprintf(pos, "%d", data[i]);
}
printf("%s\n", buf);

Here is a link to a demo on ideone.

answered Dec 13, 2012 at 1:58

Sergey Kalinichenko

729k85 gold badges1.2k silver badges1.6k bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Keith Nicholas · Accepted Answer · 2012-12-13 01:52:23Z

3

you can do something like...

char* format_uint_array(char *b, unsigned int* data, int length, char* delim, char* fmt)
    {
        int i;
        char s[50];
        b[0] = 0;
        for( i = 0; i < length; i++ )
        {
            s[0]=0;
            sprintf( s, fmt, data[i], (i<length-1)?delim : "");
            strcat(b, s);    
        }
        return b;
    }

then use it like

char buffer[128];
char formattedints[128];
sprintf("(%s)\n", format_uint_array(formattedints, array, 3, ", ", "%d%s"));

answered Dec 13, 2012 at 1:52

Keith Nicholas

44.4k15 gold badges101 silver badges168 bronze badges

1 Comment

imreal Over a year ago

Is this necessary s[0]=0;?

perreal · Accepted Answer · 2012-12-13 01:59:05Z

2

No loops:

#include <stdio.h>

int array[3] = {1, 2, 3};     
char buffer[128];

char *array_to_str(char * str, int *array, unsigned int n) {
  int r;
  if (n == 0) return 0;
  if (n == 1) r = sprintf(str, "%d", array[0]);
  else        r = sprintf(str, "%d, ", array[0]);
  array_to_str(str + r, array + 1, n - 1); 
  return str;
}

int main() { 
  printf("(%s)\n", array_to_str(buffer, array, 3));
  return 0;  
}

answered Dec 13, 2012 at 1:59

perreal

98.7k23 gold badges159 silver badges187 bronze badges

1 Comment

Chirag Over a year ago

I probably should have been more specific in my question. The format of the string may not necessarily be "(%d, %d, etc)". Although this did give me the idea of using substrings to replace one %d at a time. Thanks!

Collectives™ on Stack Overflow

sprintf and unsigned int array in C

3 Answers 3

Comments

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related