C- leading zero without printf

Question

Ethernet starter kit(PIC32MX9795F512L)
language: C
MPLAB IDE 8.92
Compiler: XC32 v1.3

Hello i want to add leading zeros to my variables. At the end i want to use the in an array. For example: c=10*a+b. When c=5 it should be 05. I cant use any printf function or am I wrong?

What are you using for "output" that isn't `printf`? Do you have `sprintf`? — Some programmer dude, Apr 01 '16 at 19:28
If you're saying that you don't think `printf()` can format decimal integers with leading zeroes, then you are mistaken. Just use the `0` flag in the relevant conversion specifier. — John Bollinger, Apr 01 '16 at 19:32
You can write one leading zero with `printf("0%d", number);` except when `number == 0` you can weed that special case out, unless you *do* want `"00"` output: one leading zero. — Weather Vane, Apr 01 '16 at 19:37
By the way, [a good `printf` (and family) reference](http://en.cppreference.com/w/c/io/fprintf) might be good to read. — Some programmer dude, Apr 01 '16 at 19:38
Possible duplicate of [Printing leading 0's in C?](http://stackoverflow.com/questions/153890/printing-leading-0s-in-c) — Joel, Apr 01 '16 at 19:45
There are various ways to *print* an integer with leading zeros. `printf` is the most obvious; if there's some reason you can't use `printf`, please explain why. But you can't add leading zeros to a *stored* integer. Integers are stored in binary, not as human-readable characters. — Keith Thompson, Apr 01 '16 at 20:10
@Joel Almost all the answers there use a form of `printf`, which he says he can't use for some reason. — Barmar, Apr 01 '16 at 20:15
@Barmar I don't think he said he is not allowed to use `printf`. He said "*I cant use any printf function or am I wrong?"* — Weather Vane, Apr 01 '16 at 21:52

score 1 · Answer 1 · answered Apr 01 '16 at 19:46

You can use printf() to simply print a formatted number to standard output:

int c = 5;
fprintf(stdout, "c [%02d]\n", c);

If you can't use printf(), another option is to store the padded value in a char * or string. You can instead use sprintf() to write the formatted string to a char * buffer.

For example:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main (int argc, char *argv[]) {
{
    char* c_str = NULL;
    int c_int = 5;
    int c_str_length = 3; /* two bytes for "0", "5", and one byte for the nul terminator */
    c_str = malloc(c_str_length); 
    if (!c_str) {
        fprintf(stderr, "Error: Could not allocate space for string!\n");
        return EXIT_FAILURE;
    }
    int n = sprintf(c_str, "%02d", c_int);
    if (n != c_str_length) { 
        fprintf(stderr, "Error: Something went wrong in writing the formatted string!\n");
        free(c_str);
        return EXIT_FAILURE;
    }
    fprintf(stdout, "c_str: [%s]\n", c_str);
    free(c_str);
    return EXIT_SUCCESS;
}

If you go this route, you can see how you could do some error checking along the way. You'll need to think about string length (hint: log10()), or use a static char [] array in place of a char * of sufficiently long length.

Weather Vane · Answer 2 · 2016-04-01T20:03:35.847

It is quite easy to add a leading zero, provided you take care of negative values too. You said you want to write to an array, so I used sprintf but if you want to output directly, you can use printf in a similar way.

char cstr[24];
int c = 10 * a + b;
if (c > 0) {
    sprintf(cstr, "0%d", c);
} else if (c < 0) {
    sprintf(cstr, "-0%d", -c);
} else {
    //sprintf(cstr, "00");                
    sprintf(cstr, "0");                // depending on your needs
}

C- leading zero without printf

2 Answers2