How to change a char to ASCII form?

Question

I need change my program, how do I use some function change a char to ASCII? my mobile receive data from 8051, always show '0', it's not true.

char to ASCII code.

void Data_TX(unsigned char Y)
{
    unsigned char Buff_Y[3];

    Buff_Y[2] = (Y / 100) + 0x30;
    Buff_Y[1] = (Y / 10) % 10+0x30;
    Buff_Y[0] = (Y % 10) + 0x30;

    SBUF = *Buff_Y;
    while (TI == 0);
    TI = 0;
}

this's my circuit original code, LCD interfacing with 8051.

bit Sensor_read(unsigned char read[5]);

use my function.

Data_TX(read[2]); //read data.
Data_TX((int)read[2]); //read data, this program can't to run.

Very unclear, what do you mean `char` to ascii? That doesn't make sense. — Iharob Al Asimi, Nov 24 '15 at 02:30
char is a variable type (in C, C++ and some other languages) to store numbers. ASCII is a method to assign symbols to numbers. What do you want? — deviantfan, Nov 24 '15 at 02:31
@deviantfan c++ data type? The OP: you should also decide what language it is, because the solution will depend on that. — Iharob Al Asimi, Nov 24 '15 at 02:32
No, just don't say it's a c++ type. It's a data type. It exists in c and c++ and other languages. — Iharob Al Asimi, Nov 24 '15 at 02:33
yes, this's C code from 8051.I need use 8051 receive data to mobile. — JDQQ8051, Nov 24 '15 at 02:45
because, my mobile need use ASCII to receive, I need change Type. — JDQQ8051, Nov 24 '15 at 02:47
@iharob ok, I know this's a data type, not a c/c++ type, There must be some misunderstanding, I'm sorry. — JDQQ8051, Nov 24 '15 at 02:53
Step by step, post some more code, and describe what are you trying to do. Then remember than division in rounded so /100 will give you '3' in case of Y>249. — Newbie, Nov 24 '15 at 02:54
@TheBook What are you talking about? Integer division truncates the remainder, so **`299 / 100` will always be `2`**. This is one of the few times it should be safe for you to *try it and see*. — autistic, Nov 24 '15 at 03:25

autistic · Accepted Answer · 2015-11-24T03:28:41.363

Buff_Y[2] = (Y / 100) + 0x30;
Buff_Y[1] = (Y / 10) % 10+0x30;
Buff_Y[0] = (Y % 10) + 0x30;

This is all good. If you print those three bytes, you will see that they contain the right values.

SBUF = *Buff_Y;

This, however, indicates confusion. *Buff_Y is equivalent to Buff_Y[0]. That is a single char value... If you can only store a single char value into SBUF, then you can't store the three char values you want to store into it.

If we use logic, we can make this inference: That char value will be '0' if the input you provide is evenly divisible by 10 (that is, when Y % 10 is 0)...

While you're thinking about how to change this line of code (and the parts of your code that we can't see), you might also want to think carefully about while (TI == 0);. There ought to be a better solution to whatever problem that is attempting to solve.

score -1 · Answer 2 · answered Nov 24 '15 at 03:43

-1

I'm change this code, can be run, but just received digit in ones.

void Data_TX(unsigned char Y)

{ unsigned char Buff_Y[3];

Buff_Y[3] = (Y / 100) + 0x30;
Buff_Y[2] = (Y%100)/10+0x30;
Buff_Y[1] = (Y / 10) % 10+0x30;
Buff_Y[0] = (Y % 10) + 0x30;

SBUF = *Buff_Y;
while (TI == 0);
TI = 0;

}

Data_TX((int)read[2]);

answered Nov 24 '15 at 03:43

JDQQ8051

3
4

`unsigned char Buff_Y[3];` denotes an array of three (not four) `unsigned char`s. They are `Buff_Y[0]`, `Buff_Y[1]` and `Buff_Y[2]`. `(Y%100)/10+0x30` is functionally identical to `(Y / 10) % 10+0x30`, so the middle character will always be repeated. Stop guessing... – autistic Nov 24 '15 at 14:00
thanks all, I'm trying to run it. this program could be run. I'm add a loop, problem is solving. – JDQQ8051 Nov 24 '15 at 15:13
Do not post answers as updates to a question. SO does not use a forum style thread, but a Q & A style. If you would like to add to your original question, use the edit link under the question and add to the previous content. (People often like to mark such additional sections in their questions with "Update:") – Toby Aug 23 '17 at 10:48

How to change a char to ASCII form?

2 Answers2