How to move uint32_t number to char[]? - c

Question

I have to copy uint32_t number into the middle of the char[] buffer.
The situation is like this:
char buf[100];
uint8_t pos = 52; // position in buffer to which I want to copy my uint32_t number
uint32_t seconds = 23456; // the actual number
I tried to use memcpy like this:
memcpy(&buf[position], &seconds, sizeof(seconds));
But in buffer I'm getting some strange characters, not the number i want
I also tried using byte-shifting
int shiftby = 32;
for (int i = 0; i < 8; i++)
{
buf[position++] = (seconds >> (shiftby -= 4)) & 0xF;
}
Is there any other option how to solve this problem?

Answer 1

What you're doing in your memcpy code is to put the value 23456 in buff, starting at byte 52 (so bytes 52-55, since the size of seconds is 4 bytes). What you want to do (if I understand you correctly) is to put the string "23456" in buff, starting at byte 52. In this second case, each character takes one byte, and each byte would hold the ASCII value of its character.
Probably the best way to do that is to use snprintf:
int snprintf(char *buffer, size_t n, const char *format-string,
argument-list);
In your example:
snprintf(&buff[position], 5, "%d", seconds)
Note that the n arguments holds the number of digits, rather than the size of the variable. As I said - you take one byte per digit/character.
Obviously you should calculate the number of digits in seconds rather than hard-code it if it can change, and you should also check the return value of snprintf to see if the operation was performed successfully

Answer 2

It is unclear how you are intending to represent this uint32_t, but your code fragment suggest that you are expecting hexadecimal (or perhaps BCD). In that case:
for( int shiftby = 28; shiftby >= 0 ; shiftby -= 4 )
{
char hexdigit = (seconds >> shiftby) & 0xF ;
buf[position++] = hexdigit < 10 ? hexdigit + '0' : hexdigit + 'A' - 10 ;
}
Note that the only real difference between this and your code is the conversion to hex-digit characters by adding conditionally either '0' or 'A' - 10. The use of shiftby as the loop control variable is just a simplification or your algorithm.
The issue with your code is that it inserted integer values 0 to 15 into buf and the characters associated with these values are all ASCII control characters, nominally non-printing. How or whether they render as a glyph on any particular display depends on what you are using to present them. In Windows console for example, printing characters 0 to 15 results in the following:
00 = <no glyph>
01 = '☺'
02 = '☻'
03 = '♥'
04 = '♦'
05 = '♣'
06 = '♠'
07 = <bell> (emits a sound, no glyph)
08 = <backspace>
09 = <tab>
10 = <linefeed>
11 = '♂'
12 = '♀'
13 = <carriage return>
14 = '♫'
15 = '☼'
The change above transforms the values 0 to 15 to ASCII '0'-'9' or 'A'-'F'.
If a hexadecimal presentation is not what you were intending then you need to clarify the question.
Note that if the encoding is BCD (Binary Coded Decimal) where each decimal digit is coded into a 4 bit nibble, then the conversion can be simplified because the range of values is reduced to 0 to 9:
char bcddigit = (seconds >> shiftby) & 0xF ;
buf[position++] = bcddigit + '0' ;
but the hex conversion will work for BCD also.