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sprintf buffer

 
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vtrx



Joined: 11 Oct 2017
Posts: 130

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sprintf buffer
PostPosted: Fri May 06, 2022 6:14 pm     Reply with quote

I'm trying to use 'sprintf' to load a buffer with decimal numbers to show in the 7 segment module (TM1637) but I don't understand how the buffer is being loaded.
The function that displays the digits searches the table for the numbers that correspond to their decimal value.
It's 4 digits.

Code:

const char segmentMap[] = {
    0x3f, 0x06, 0x5b, 0x4f, 0x66, 0x6d, 0x7d, 0x07, // 0-7
    0x7f, 0x6f, 0x77, 0x7c, 0x39, 0x5e, 0x79, 0x71, // 8-9, A-F
    0x00
};
char         digitos[8];
int            sec, h, m, s;
...
void Mostra_digitos(void)
 {
    h = (sec/3600);   
    m = (sec -(3600*h))/60;   
    s = (sec -(3600*h)-(m*60));
    sprintf(digitos,"%02d%02d%02d",h,m,s);
    start();
    writeByte(0x40);
    ask();
    Stop();
    Start();
    writeByte(0xc0);
    ask();
        writeByte(segmentMap[digitos[0]]);//m
        ask();
        writeByte(segmentMap[digitos[1]]);//m
        ask();       
        writeByte(segmentMap[digitos[2]]);//s
        ask();
        writeByte(segmentMap[digitos[3]]);//s
        ask();       
    Stop();
 }
//---


Code:
   
...
   sec = 360; 
   Mostra_digitos();


This code snippet converts seconds into hours, and returns the correct value if using a terminal or simulator.
Code:
    h = (sec/3600);   
    m = (sec -(3600*h))/60;   
    s = (sec -(3600*h)-(m*60));
    sprintf(digitos,"%02d%02d%02d",h,m,s);
...


sec=360 returns 000600
I tried using 'sprintf' to load the buffer with digits[0]=0,digits[1]=0,digits[2]=0,digits[3]=6, digits[4]=0 and digits[5]=0 but 'sprintf' didn't load the buffer this way.
How could I do?
PCM programmer



Joined: 06 Sep 2003
Posts: 21647

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Re: sprintf buffer
PostPosted: Fri May 06, 2022 8:19 pm     Reply with quote

vtrx wrote:

int sec, h, m, s;
sec = 360;

From a previous topic, you are using the 18F4550.
In the PCH compiler, an 'int' is an unsigned 8-bit number.
It can only go up to 255. You are trying to put 360 into an int.
This will not work.
PrinceNai



Joined: 31 Oct 2016
Posts: 316
Location: Montenegro

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PostPosted: Fri May 06, 2022 8:26 pm     Reply with quote

Was just writing this. Your declarations are not ok.
Quote:

int sec, h, m, s;

h = (sec/3600);
m = (sec -(3600*h))/60;
s = (sec -(3600*h)-(m*60));


This can't work, since you declared at least sec as a wrong type (a number between 0 and 255), than divided that by 3600 and expected to get an integer bigger than 0 from that division. Try it in Excel with the result cell formatted as a number without decimals. Always 0. sec should be int16
vtrx



Joined: 11 Oct 2017
Posts: 130

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PostPosted: Fri May 06, 2022 8:57 pm     Reply with quote

You are really correct!
I tested the conversion code using this page.
https://www.programiz.com/c-programming/online-compiler/

Code:
int sec, h, m, s;

   printf("Input seconds: ");
   scanf("%d", &sec);   
   h = (sec/3600);    
   m = (sec -(3600*h))/60;   
   s = (sec -(3600*h)-(m*60));   
   printf("H:M:S - %02d:%02d:%02d\n",h,m,s);   
   return 0;

Is it necessary for h,m and s to be int16 too?
I just changed sec to int16 and it worked.
thank you all.
PrinceNai



Joined: 31 Oct 2016
Posts: 316
Location: Montenegro

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PostPosted: Fri May 06, 2022 9:23 pm     Reply with quote

If you know the size of the result, it isn't or shouldn't be necessary. If not, you lose 3 bytes of RAM and some extra processor time by declaring them all as int16. From personal experience it is better to work with the same size of the variables.
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