from int to char, Sending byte by byte

Question

I was wondering how can I convert a 6 digit number for example 198200, to a char for example Tx = ['1', '9', '8', '2', '0', '0'], so that later I can for example write:

*p_tx_buffer++ = Tx[2];

And then I will only send a the '8'.

See http://stackoverflow.com/questions/8257714/how-to-convert-an-int-to-string-in-c — AdamSkwersky, Dec 01 '15 at 21:44
I believe http://stackoverflow.com/questions/5590381/easiest-way-to-convert-int-to-string-in-c is a much more complete dupe. — vsoftco, Dec 01 '15 at 21:46
@vsoftco Note: suggested dupe is for `C++` and this post is ill tagged `C/C++` post. — chux - Reinstate Monica, Dec 01 '15 at 21:48

score 1 · Answer 1 · answered Dec 01 '15 at 21:42

1

A simple quick solution is

char array[100];
int number = 198200;
if (snprintf(array, sizeof(array), "%d", number) >= sizeof(array))
    fprintf(stderr, "there is not enough room for the string\n");
else
    fprintf(stdout, "array[2] = %c\n", array[2]);

answered Dec 01 '15 at 21:42

Iharob Al Asimi

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Thank you for your answer, but if I want to get only the saved char, like only Tx, because i want to use it later, what should I consider ? because with the code you have provided i believe i can only print it ,i want it as a saved variable? Thank you – Mehdi Dec 01 '15 at 22:11
What do you think `array` is? – Iharob Al Asimi Dec 01 '15 at 22:45

from int to char, Sending byte by byte

1 Answers1