Int to Char help C, STM32F429ZI

[Shahin]

I'm creating an int to char function, but struggling with parameter passing

 

char num_to_char(short Number, int POS) // Less than 100,000
{
char Num_char[5] = {'0','0','0','0','0'};
    
    while (Number > 10000)
    {
    Num_char[0]++;
    Number = Number - 10000;    
    }

    while (Number > 1000)
    {
    Num_char[1]++;
    Number = Number - 1000;    
    }
    
    while (Number > 100)
    {
    Num_char[2]++;
    Number = Number - 100;    
    }

    while (Number > 10)
    {
    Num_char[3]++;
    Number = Number - 10;    
    }
    
    while (Number > 1)
    {
    Num_char[4]++;
    Number = Number - 1;    
    }
return Num_char[POS];

 

 

 

 

calling back

voltage[0] = num_to_char( Number,  0);
voltage[1] = '.';
voltage[2] = num_to_char( Number, 1 );
voltage[3] = num_to_char( Number,  2);
voltage[4] = num_to_char( Number,  3);
voltage[5] =num_to_char( Number,  4);
voltage[6] = 'V';        

 

Any help?

[Shahin]

Solved, Not sure how to delete post

[reniat]

43 minutes ago, Shahin said:

Solved, Not sure how to delete post

even better than deleting it, you could post your solution so that the next person who finds this while googling can benefit from it

[wanderingfool2]

On 12/19/2018 at 12:26 PM, reniat said:

even better than deleting it, you could post your solution so that the next person who finds this while googling can benefit from it

 

My assumption would be that @Shahin realized the missing = signs in while loop (just a guess though) or realizing a short isn't good to use in this case.

 

Boundary cases are important

while(num >= 10000) captures the case where the number is 10000.

 

Anyways, @Shahin it is a good attempt, and glad you figured out a solution.  I do compliment you no the '0' and adding though.  A note, shorts range from -33k - 33k (roughly), so an int might be better (and then do an if statement to check if it is above 100k).  Also, since you are only returning 1 number, you can simplify it more with just some integer math

 

e.g.

return '0' + (Number / (int)pow(10,POS))%10; //It's late, I might be wrong in syntax, logic and such
//The concept being you divide by 10000 or whatever, and then %10 gets you the remainder (the value).

 

[Mira Yurizaki]

Well, because I'm bored, here's a solution in C if anyone cares:

Spoiler

#define MAX_SIZE 11
static char int_string[MAX_SIZE] = {0}; /* Assume 32-bit int, max digits is 10 + 1 for NUL */

void int_to_char(int input_num){
  	int exp_val = 1000000000;
  	int place = 0;
  
    /* Special case: if input is 0, fill in 0 and exit immediately*/
    if (input_num == 0){
      	int_string[0] = '0';
    }
  
  	/* Find out where the number starts from the most significant end */
    while ((input_num / exp_val) == 0){
      	exp_val /= 10;
    }
  
  	/* Fill in the numbers. */
    while (exp_val > 0){
      int_string[place++] = (char)((input_num / exp_val) + 0x30); /* 0x30 is 0 in most character encodings */
      input_num %= exp_val;
      exp_val /= 10;
    }
}

Notes:

• This function either has two constant amount of comparisons, either 1 or 12. However it should be assumed 12 is going to always be the case.
• This approach uses most significant to least significant digit to avoid adding leading zeroes. Least to most significant digit approach can't tell where the last actual most significant digit is until it traverses the whole number.

 

[0x21]

#include <stdio.h>
  
char int_to_char(int n)
{
  char number[11];
  sprintf(number, "%d", n);
  return *number;
}

 

[Unimportant]

2 hours ago, mshaugh said:

#include <stdio.h>
  
char int_to_char(int n)
{
  char number[11];
  sprintf(number, "%d", n);
  return *number;
}

 

That only returns a single character, not a string.

 

[0x21]

3 hours ago, Unimportant said:

That only returns a single character, not a string.

You're definitely not wrong.

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
  
char* int_to_char(int n)
{
  int length = (int)(n ? log10(n) + 1 : 1);
  char* number = (char*)(malloc(length+1));
  sprintf(number, "%d", n);
  
  return number;
}

int main(void)
{
  int n = 10000;
  
  char *number;
  number = int_to_char(n);
  
  printf("%s\n", number);
  free(number);
  
  return 0;
}

Alternatively just use `sprintf` in place of your own function.

[Mira Yurizaki]

On 12/22/2018 at 3:36 AM, mshaugh said:

You're definitely not wrong.

Alternatively just use `sprintf` in place of your own function.

While I like the sprintf method for being concise, I do have one particular issue with it. OP's target is an embedded systems processor.

 

Why is this an issue? Getting dynamic memory allocation to work on embedded systems processors seems to be a pain. A quick glance online about getting it setup involved things like messing with the linker script, memory mapping, among other things. And then there's the design aspect of it, such as figuring out a good enough heap size, if you have any way of gracefully handling memory leaks, etc. It also makes the system less deterministic. For all those reasons, especially when you're coding to bare metal, I don't recommend any sort of dynamic memory system for this sort of target.

Edited December 23, 2018 by M.Yurizaki