How to Convert from Int to Char*

How to convert integer to char

Use Char(n + '0'). This will add the ASCII offset of the 0 digit and fix the rest of the digits too. For example:

julia> a = 5
5

julia> Char(a+'0')
'5': ASCII/Unicode U+0035 (category Nd: Number, decimal digit)

Also note, timing with @time is a bit problematic, especially for very small operations. It is better to use @btime or @benchmark from BenchmarkTools.jl .

C - casting int to char and append char to char

You can use itoa function to convert the integer to a string.

You can use strcat function to append characters in a string at the end of another string.

If you want to convert a integer to a character, just do the following -

int a = 65;
char c = (char) a;

Note that since characters are smaller in size than integer, this casting may cause a loss of data. It's better to declare the character variable as unsigned in this case (though you may still lose data).

To do a light reading about type conversion, go here.

If you are still having trouble, comment on this answer.

Edit

Go here for a more suitable example of joining characters.

Also some more useful link is given below -

1. http://www.cplusplus.com/reference/clibrary/cstring/strncat/
2. http://www.cplusplus.com/reference/clibrary/cstring/strcat/

Second Edit

char msg[200];
int msgLength;
char rankString[200];

........... // Your message has arrived
msgLength = strlen(msg);
itoa(rank, rankString, 10); // I have assumed rank is the integer variable containing the rank id

strncat( msg, rankString, (200 - msgLength) );  // msg now contains previous msg + id

// You may loose some portion of id if message length + id string length is greater than 200

Third Edit

Go to this link. Here you will find an implementation of itoa. Use that instead.

how to convert from int to char*?

• In C++17, use std::to_chars as:

  std::array<char, 10> str;
  std::to_chars(str.data(), str.data() + str.size(), 42);
  

• In C++11, use std::to_string as:

  std::string s = std::to_string(number);
  char const *pchar = s.c_str();  //use char const* as target type
  

• And in C++03, what you're doing is just fine, except use const as:

  char const* pchar = temp_str.c_str(); //dont use cast
  

Converting an int to a char in C?

Given the revised requirement to get digit '0' into string[i] if number == 0, and similarly for values of number between 1 and 9, then you need to add '0' to the number:

assert(number >= 0 && number <= 9);
string[i] = number + '0';

The converse transform is used to convert a digit character back to the corresponding number:

assert(isdigit(c));

int value = c - '0';

Convert from Int to char - string - double - int

128 isn't the value of a printable character. You're getting ? because the display wants to show you that there is a Char value there but it can't show you the character that the value represents.

Since it is a character (even if it is unshowable), it can be turned into a String of length 1. But that String cannot be turned into a number because only number representations (like "3.56" or "-79") can be converted to a number.

scala> 128.toChar.toString.toDouble.toInt
java.lang.NumberFormatException: For input string: ""
  at java.base/jdk.internal.math.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:2054)
  at java.base/jdk.internal.math.FloatingDecimal.parseDouble(FloatingDecimal.java:110)
  at java.base/java.lang.Double.parseDouble(Double.java:543)
  at scala.collection.immutable.StringLike.toDouble(StringLike.scala:317)
  at scala.collection.immutable.StringLike.toDouble$(StringLike.scala:317)
  at scala.collection.immutable.StringOps.toDouble(StringOps.scala:29)
  ... 28 elided

scala> 55.toChar.toString.toDouble.toInt
res33: Int = 7

In order to restore the original value it has to be transitioned to a type with a different toString method.

scala> 128.toChar.toShort.toString.toDouble.toInt
res34: Int = 128

Converting an integer to char by adding '0' - what is happening?

Each symbol has an numeric representation, so basically each char is a number. Here is the table of characters and its values.

_{(source: asciitable.com)}

So, in your code name[1] = numbers[1];, you assign name[1] to 2, which is equal to symbol STX in the ASCII table above, which is not a printable character, thats why you get that incorrect output.

When you add '0' to your name[1], it is equal to add 48 to 2, so the symbol with the number 50 is '2', that's why you get that correct output.

C#: Convert int to char, but in literal not unicode representation

If I get you correctly, this is what you want

int i1 = 3;
char c = (char)(i1 + '0');

Console.WriteLine(c);

'0' is 48 in ASCII. Adding 3 to 48 is 51 and converting 51 to char is 3.

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