98

I have 

int i = 6;

and I want 

char c = '6'

by conversion. Any simple way to suggest?

EDIT: also i need to generate a random number, and convert to a char, then add a '.txt' and access it in an ifstream.

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11 Answers 11

Reset to default
417

Straightforward way:

char digits[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9' }; char aChar = digits[i];

Safer way:

char aChar = '0' + i;

Generic way:

itoa(i, ...)

Handy way: 

sprintf(myString, "%d", i)

C++ way: (taken from Dave18 answer)

std::ostringstream oss; oss << 6;

Boss way:

Joe, write me an int to char converter

Studboss way:

char aChar = '6';

Joe's way:

char aChar = '6'; //int i = 6;

Nasa's way:

//Waiting for reply from satellite...

Alien's way: '9'

//Greetings.

God's way:

Bruh I built this

Peter Pan's way:

char aChar; switch (i) { case 0: aChar = '0'; break; case 1: aChar = '1'; break; case 2: aChar = '2'; break; case 3: aChar = '3'; break; case 4: aChar = '4'; break; case 5: aChar = '5'; break; case 6: aChar = '6'; break; case 7: aChar = '7'; break; case 8: aChar = '8'; break; case 9: aChar = '9'; break; default: aChar = '?'; break; }

Santa Claus's way:

//Wait till Christmas! sleep(457347347);

Gravity's way:

//What

'6' (Jersey) Mikes'™ way:

//

SO way:

Guys, how do I avoid reading beginner's guide to C++?

My way:

or the highway.

Comment: I've added Handy way and C++ way (to have a complete collection) and I'm saving this as a wiki.

Edit: satisfied?

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13 Comments

You meant "boss" as in superior, but I read it as in "boss", as in "Bruce Springsteen is the boss." Maybe you could revise for that case?
About about Joe's way? Or Nasa's way? or Aliens way? Or Gods way? Or Peter Pans way? How about Santa Claus's way? Or Gravity's way? You should start a list? Maybe check it twice just in case you missed something?
@AbstractDissonance can a get that with Mikes Way™ ... I like me some Jersey Mikes.
This made me laugh so hard.
This has to be the best answer on SO xD
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38

This will only work for int-digits 0-9, but your question seems to suggest that might be enough.

It works by adding the ASCII value of char '0' to the integer digit.

int i=6; char c = '0'+i; // now c is '6'

For example:

'0'+0 = '0' '0'+1 = '1' '0'+2 = '2' '0'+3 = '3'

Edit

It is unclear what you mean, "work for alphabets"? If you want the 5th letter of the alphabet:

int i=5; char c = 'A'-1 + i; // c is now 'E', the 5th letter.

Note that because in C/Ascii, A is considered the 0th letter of the alphabet, I do a minus-1 to compensate for the normally understood meaning of 5th letter.

Adjust as appropriate for your specific situation.
(and test-test-test! any code you write)

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2 Comments

Confusing when choosing between small and capital letters.
Well, you need to tell us what you want. "Work for alphabets" really isn't very descriptive.
16

Just FYI, if you want more than single digit numbers you can use sprintf:

char txt[16]; int myNum = 20; sprintf(txt, "%d", myNum);

Then the first digit is in a char at txt[0], and so on.

(This is the C approach, not the C++ approach. The C++ way would be to use stringstreams.)

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3 Comments

Always prefer snprintf over sprintf (consider a hypothetical system with 128 bit ints or whatever). +1 anyway for an approach that won't break in character sets where the numbers aren't consecutive.
@Mark B: Standard C requires that the numbers be consecutive.
@Mark Right, that's definitely safer from a buffer-overflow perspective.
5

This is how I converted a number to an ASCII code
0 though 9 in hex code is 0x30 - 0x39
6 would be 0x36

unsigned int temp = 6; // or you can use 'unsigned char temp = 6; unsigned char num; num = 0x30 | temp;

This will give you the ASCII value for 6. You do the same for 0 - 9

To convert ASCII to a numeric value I came up with this code

unsigned char num, code; code = 0x39; // ASCII Code for 9 in Hex num = 0 & 0 F & code;
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4

My way to do this job is:

char to int char var; cout<<(int)var-48; int to char int var; cout<<(char)(var|48);

And I write these functions for conversions:

int char2int(char *szBroj){ int counter=0; int results=0; while(1){ if(szBroj[counter]=='\0'){ break; }else{ results*=10; results+=(int)szBroj[counter]-48; counter++; } } return results; } char * int2char(int iNumber){ int iNumbersCount=0; int iTmpNum=iNumber; while(iTmpNum){ iTmpNum/=10; iNumbersCount++; } char *buffer=new char[iNumbersCount+1]; for(int i=iNumbersCount-1;i>=0;i--){ buffer[i]=(char)((iNumber%10)|48); iNumber/=10; } buffer[iNumbersCount]='\0'; return buffer; }
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2

itoa()

http://www.cplusplus.com/reference/clibrary/cstdlib/itoa/

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8 Comments

I like this approach better than the character arithmetic suggested by abelenky because it doesn't depend on working with the ASCII character set (even though that is a fairly safe assumption for many applications). However, itoa stores its result in a string, so you need to provide a character array and extract the character from it. Your answer would be more helpful to neophytes if you edited it to show those extra steps.
-1 both for itoa (nonstandard MS function) and for linking to perhaps the worst site for C and C++ [mis]information.
"standard" way in C++ would be using streams, all the other methods mentioned so far have some problems.
@R..: sorry, could you link the best site for c and c++ then ?
well, then its inferior compared to cplusplus.com
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2

I suppose that

std::to_string(i)

could do the job, it's an overloaded function, it could be any numeric type such as int, double or float

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1 Comment

for example, char c = to_string(9)[0];
0

"I have int i = 6; and I want char c = '6' by conversion. Any simple way to suggest?"

There are only 10 numbers. So write a function that takes an int from 0-9 and returns the ascii code. Just look it up in an ascii table and write a function with ifs or a select case.

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0

Alternative way, But non-standard.

int i = 6; char c[2]; char *str = NULL; if (_itoa_s(i, c, 2, 10) == 0) str = c;

Or Using standard c++ stringstream

 std::ostringstream oss; oss << 6;
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3 Comments

itoa is non-standard and unsafe (could cause buffer overrun).
Okay, you're edited it to the _s version now. Still non-standard, but I removed my downvote!
Right! I'm mistyped the object name which should be 'oss' :)
0

Doing college work I gathered the data I found and gave me this result:

"The input consists of a single line with multiple integers, separated by a blank space. The end of the entry is identified by the number -1, which should not be processed."

#include <iostream> #include <cstdlib> using namespace std; int main() { char numeros[100]; //vetor para armazenar a entrada dos numeros a serem convertidos int count = 0, soma = 0; cin.getline(numeros, 100); system("cls"); // limpa a tela for(int i = 0; i < 100; i++) { if (numeros[i] == '-') // condicao de existencia do for i = 100; else { if(numeros[i] == ' ') // condicao que ao encontrar um espaco manda o resultado dos dados lidos e zera a contagem { if(count == 2) // se contegem for 2 divide por 10 por nao ter casa da centena soma = soma / 10; if(count == 1) // se contagem for 1 divide por 100 por nao ter casa da dezena soma = soma / 100; cout << (char)soma; // saida das letras do codigo ascii count = 0; } else { count ++; // contagem aumenta para saber se o numero esta na centena/dezena ou unitaria if(count == 1) soma = ('0' - numeros[i]) * -100; // a ideia é que o resultado de '0' - 'x' = -x (um numero inteiro) if(count == 2) soma = soma + ('0' - numeros[i]) * -10; // todos multiplicam por -1 para retornar um valor positivo if(count == 3) soma = soma + ('0' - numeros[i]) * -1; /* caso pense em entrada de valores na casa do milhar, deve-se alterar esses 3 if´s alem de adicionar mais um para a casa do milhar. */ } } } return 0; }

The comments are in Portuguese but I think you should understand. Any questions send me a message on linkedin: https://www.linkedin.com/in/marcosfreitasds/

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1 Comment

It would be really great if you, or somebody else of course, could translate the comments!
-4 
 A PROGRAM TO CONVERT INT INTO ASCII. #include<stdio.h> #include<string.h> #include<conio.h> char data[1000]= {' '}; /*thing in the bracket is optional*/ char data1[1000]={' '}; int val, a; char varray [9]; void binary (int digit) { if(digit==0) val=48; if(digit==1) val=49; if(digit==2) val=50; if(digit==3) val=51; if(digit==4) val=52; if(digit==5) val=53; if(digit==6) val=54; if(digit==7) val=55; if(digit==8) val=56; if(digit==9) val=57; a=0; while(val!=0) { if(val%2==0) { varray[a]= '0'; } else varray[a]='1'; val=val/2; a++; } while(a!=7) { varray[a]='0'; a++; } varray [8] = NULL; strrev (varray); strcpy (data1,varray); strcat (data1,data); strcpy (data,data1); } void main() { int num; clrscr(); printf("enter number\n"); scanf("%d",&num); if(num==0) binary(0); else while(num>0) { binary(num%10); num=num/10; } puts(data); getch(); }

I check my coding and its working good.let me know if its helpful.thanks.

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1 Comment

you could work on your indentation... and think a lot about other problems in that code.

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