C++ 将 int 转换为 ASCII 字符
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Convert an int to ASCII character
提问by user963241
I have
我有
int i = 6;
and I want
而且我要
char c = '6'
by conversion. Any simple way to suggest?
通过转换。有什么简单的建议方法吗？
EDIT:also i need to generate a random number, and convert to a char, then add a '.txt' and access it in an ifstream.
编辑：我还需要生成一个随机数，并转换为一个字符，然后添加一个 '.txt' 并在 ifstream 中访问它。
回答by Eugene Mayevski 'Callback
Straightforward way:
直截了当的方式：
char digits[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9' };
char aChar = digits[i];
Safer way:
更安全的方法：
char aChar = '0' + i;
Generic way:
通用方式：
itoa(i, ...)
Handy way:
方便的方法：
sprintf(myString, "%d", i)
C++ way:(taken from Dave18 answer)
C++ 方式：（取自 Dave18 的回答）
std::ostringstream oss;
oss << 6;
Boss way:
老板方式：
Joe, write me an int to char converter
乔，给我写一个 int 到 char 转换器
Studboss way:
站长方式：
char aChar = '6';
字符 aChar = '6';
Joe's way:
乔的方式：
char aChar = '6'; //int i = 6;
字符 aChar = '6'; //int i = 6;
Nasa's way:
美国宇航局的方法：
//Waiting for reply from satellite...
//等待卫星回复...
Alien's way: '9'
外星人的方式：'9'
//Greetings.
//你好。
God's way:
大神之道：
Bruh I built this
Bruh 我建了这个
Peter Pan's way:
彼得潘的方法：
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:
圣诞老人的方式：
//Wait till Christmas!
sleep(457347347);
Gravity's way:
重力方式：
//What
//什么
'6' (Jersey) Mikes'? way:
'6'（球衣）麦克斯'？道路：
//
//
SO way:
所以方式：
Guys, how do I avoid reading beginner's guide to C++?
伙计们，我如何避免阅读 C++ 初学者指南？
My way:
我的方式：
or the highway.
或高速公路。
Comment: I've added Handy way and C++ way (to have a complete collection) and I'm saving this as a wiki.
评论：我添加了 Handy 方式和 C++ 方式（以获得完整的集合），并将其保存为 wiki。
Edit: satisfied?
编辑：满意？
回答by abelenky
This will only work for int-digits 0-9, but your question seems to suggest that might be enough.
这仅适用于整数 0-9，但您的问题似乎表明这可能就足够了。
It works by adding the ASCII value of char '0'to the integer digit.
它的工作原理是将 char 的 ASCII 值添加'0'到整数数字上。
int i=6;
char c = '0'+i; // now c is '6'
For example:
例如：
'0'+0 = '0'
'0'+1 = '1'
'0'+2 = '2'
'0'+3 = '3'
Edit
编辑
It is unclear what you mean, "work for alphabets"? If you want the 5th letter of the alphabet:
不清楚你的意思是“为字母工作”？如果您想要字母表的第 5 个字母：
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 letterof the alphabet, I do a minus-1 to compensate for the normally understood meaning of 5th letter.
请注意，因为在 C/Ascii 中，A 被认为是字母表的第 0 个字母，所以我做了一个减 1 来补偿通常理解的第 5 个字母的含义。
Adjust as appropriate for your specific situation.
(and test-test-test!any code you write)
根据您的具体情况进行适当调整。
（和测试测试测试！你写的任何代码）
回答by Nathan S.
Just FYI, if you want more than single digit numbers you can use sprintf:
仅供参考，如果您想要多于一位的数字，您可以使用 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.
然后第一个数字在 txt[0] 处的字符中，依此类推。
(This is the C approach, not the C++ approach. The C++ way would be to use stringstreams.)
（这是 C 方法，而不是 C++ 方法。C++ 方法是使用字符串流。）
回答by aleksandar kamenjasevic
My way to do this job is :
我做这项工作的方法是：
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
我编写了这些转换函数
int char2int(char *szBroj){
int counter=0;
int results=0;
while(1){
if(szBroj[counter]=='unsigned int temp = 6;
or you can use unsigned char temp = 6;
unsigned char num;
num = 0x30| temp;
'){
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]='unsigned char num,code;
code = 0x39; // ASCII Code for 9 in Hex
num = 0&0F & code;
';
return buffer;
}
回答by Dorato
This is how I converted a number to an ASCII code. 0 though 9 in hex code is 0x30-0x39. 6 would be 0x36.
这就是我将数字转换为 ASCII 代码的方式。十六进制代码中的 0 到 9 是 0x30-0x39。6 将是 0x36。
int i = 6;
char c[2];
char *str = NULL;
if (_itoa_s(i, c, 2, 10) == 0)
str = c;
this will give you the ASCII value for 6. You do the same for 0 - 9
这将为您提供 6 的 ASCII 值。您对 0 - 9 执行相同操作
to convert ASCII to a numeric value I came up with this code.
将 ASCII 转换为数值我想出了这个代码。
std::ostringstream oss;
oss << 6;
回答by Newbie
回答by cpx
Alternative way, But non-standard.
替代方式，但非标准。
std::to_string(i)
Or Using standard c++ stringstream
或使用标准 C++ stringstream
#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;
}
回答by jun
"I have int i = 6; and I want char c = '6' by conversion. Any simple way to suggest?"
“我有 int i = 6; 我想通过转换得到 char c = '6'。有什么简单的建议方法吗？”
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.
只有10个数字。所以编写一个函数，它从 0-9 中获取一个 int 并返回 ascii 代码。只需在 ascii 表中查找并使用 ifs 或 select case 编写一个函数。
回答by iceSea
I suppose that
我想
##代码##could do the job, it's an overloaded function, it could be any numeric type such as int, double or float
可以完成这项工作，它是一个重载函数，它可以是任何数字类型，例如 int、double 或 float
回答by Marcos Freitas Brazil
Doing college work I gathered the data I found and gave me this result:
在做大学工作时，我收集了我发现的数据并给出了以下结果：
"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."
“输入由一行多个整数组成，由空格分隔。条目的末尾由数字 -1 标识，不应处理。”
##代码##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/
评论是葡萄牙语，但我想你应该明白。任何问题都可以在linkedin上给我发消息：https: //www.linkedin.com/in/marcosfreitasds/
