Converting an int to String in C++

Question

I have been working on a project where I need to convert an int8_t variable to a string. I did a bit of reading on this and came to a conclusion that using Strint(int8_t) is an option.

Here is my code:

int8_t matt = 0;
matt += 1;
char string[3]=String(matt);
tft.textWrite(string);// this is used to display text on an lcd display (arduino)

For some reason the method I used did not work so I researched a bit more and found out String(matt) is actually a String object. I don't really know if this is true for a fact. In order for the tft.textWrite(); to work it should look something like this.

Here is the code:

 char string[15]= "Hello, World! ";
 tft.textWrite(string);

I tried using this also:

char string[3];
sprintf(string, "%ld", matt);

However this did not work.

Can anybody help?

Thanks

`char string[3]` should be `char string[5]` as it needs to hold e.g. `-100`, i.e. 4 chars and a termination. Also: What do mean by `However this did not work` ? In which way did it fail? — Support Ukraine, Sep 06 '16 at 16:42
You can just convert back to a char array if you want to use the textWrite function, something like char string[1024]; std::strcpy(string, std::to_string(0).c_str()); — George, Sep 06 '16 at 16:43
Because what happened is it starting printing weird characters for some reason on the screen. If I just say char string[3]="1" everything prints fine. — , Sep 06 '16 at 16:44
The "duplicate" answer does not tag arduino, so you should check the avr-libc itoa(int val, char* target, int radix) solution there. Less overhead than snprintf or String(int val). — datafiddler, Sep 06 '16 at 20:41

score 0 · Answer 1 · answered Sep 06 '16 at 16:43

0

It should work fine.

sprintf(string, "%ld", matt);

%ld is used for long int. you are using int8_t which is of range -128 to +128, it is better to use %d.

answered Sep 06 '16 at 16:43

Rajmani Arya

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7

score 0 · Answer 2 · answered Sep 06 '16 at 16:47

0

This code is nearly right

char string[3];
sprintf(string, "%ld", matt);

first

"%ld"

expects a long int as a type you need to find one that can print an 8 bit int. I think its "%hh".

second your

char string[3];

is too short! what if matt is 100 then you use 3 chars but you still need room for the ending zero (and sign). So it should be at least

char string[5];

third the code

sprintf(string, "%ld", matt);

is unsafe as it can write beyond the end of string as it would in your original code, use

snprintf(string, sizeof(string), "%ld", matt);

answered Sep 06 '16 at 16:47

Surt

15,501
3
23
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As everything with rank less than `int` is promoted to int when passed to variadic function, you need to use `%d` format specifier. – Revolver_Ocelot Sep 06 '16 at 16:54
@Revolver_Ocelot, that would simplify it a bit. – Surt Sep 06 '16 at 16:58

Maxime Trimolet · Answer 3 · 2016-09-06T19:34:51.930

0

First you have to keep in mind that there is no String type in C, only fixed sized char arrays (type: char* or char[]). We still call them strings, but keep that in mind.

Your attempt with sprintf looked good to me. Are you shure your environment includes the full C Standard Library ? If not, you will have to code your own format function. If yes, have you included string.h ?

Anyway, try to use snprintf() to precise the size of the target, it is safer.

EDIT: This post applies to the good ol' C style, which I did not see was off topic

edited Sep 06 '16 at 19:34

answered Sep 06 '16 at 16:48

Maxime Trimolet

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The question is about C++, not C. C++ definitely has a string type. We don't bully ourselves with character arrays any more. – Lightness Races in Orbit Sep 06 '16 at 16:52
Please tell me you didn't believe that X) – Maxime Trimolet Sep 06 '16 at 17:36
Didn't believe what? – Lightness Races in Orbit Sep 06 '16 at 18:55
"we don't bully ourselves with character arrays anymore" Have you heard of low level or embedded development ? – Maxime Trimolet Sep 06 '16 at 19:21
Heard of it? I do it for a living, mate. None of this is relevant, though: the simple fact is that your answer is geared towards a C developer (or, fine, a C++ developer who is unable to use the C++ standard library because he is programming hardware from the 1980s), which the OP is not. – Lightness Races in Orbit Sep 06 '16 at 19:21
In fairness I did just spot the [tag:arduino] tag. If Arduino does not support the C++ standard library and/or heap allocations, please adjust your answer so that it does not talk about "C" but about the limitations of C++ on that platform. Otherwise, what I said above still applies. – Lightness Races in Orbit Sep 06 '16 at 19:27
So do I, so tell me how you avoid using it ? Isn't "dynamic" (better say undefined) behavior a capital waste of runtime as well as memory ? – Maxime Trimolet Sep 06 '16 at 19:27
You seem to be engaging in some sort of holy war over coding style. That's not what answers (or comments!) are for. Please stick to the topic at hand. – Lightness Races in Orbit Sep 06 '16 at 19:28
I am genuinely interested in talking with you in fact :D – Maxime Trimolet Sep 06 '16 at 19:28

score -1 · Answer 4 · answered Sep 06 '16 at 16:46

-1

the following code should work:

// Example program
#include <iostream>
#include <string>

int main()
{
  int8_t matt = 0;
  matt += 1;
  char string[1]; 
  sprintf(string, "%ld", matt);
  std::cout << string << std::endl;
  return 0;
}

it prints "1".

answered Sep 06 '16 at 16:46

Ivan Ustinov

49
4

you saw the "arduino" tag? Among others, there's no std::cout ;) – datafiddler Sep 06 '16 at 20:31

Converting an int to String in C++

4 Answers4