将 int 转换为 ASCII 字符

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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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itoa()

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

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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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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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最佳答案

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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"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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          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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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&0F & code;

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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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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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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/