ordinal value

just wondering, how do I get the ordinal value for an integer in C++.  I basically just moved from Delphi to C++.  I know it's for example, Ord(i) in Delphi, to get the ASCII value of an integer. (if I remember correctly)

how do I get the ASCII value in C++?  (Using C++ Builder 5)

Thanks for the help
LmoenAsked:
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TriskelionConnect With a Mentor Commented:
In C and C++, many of the simple types ARE the ordinal values depeding on how you treat them.

For instance, if you
printf("%d", '0'); // decimal
you will get it's ASCII value.

if you
printf("%c", 48);//character
you will get the character representation of the number.
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TriskelionCommented:
better example

printf("%c %d", '0','0');
will print
   0 48
(in an ASCII evironment)
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AxterCommented:
Example:

int var = 65; //65 equals 'A';
char SomeStr[5]= "";
SomeStr[0] = var;

or

char SomeStr[] = "Hello World";
int ASC_Value = SomeStr[0];//makes it equal to 'H'
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AxterCommented:
The following is an example on how to convert a number to a std::string

#include <sstream>
#include <string>

template <typename T>
std::string Ttos(T arg)
{
   std::ostringstream buffer;
   buffer << arg; // send to the ostringstream
   return buffer.str(); // capture the string
}
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AxterCommented:
Some compilers support a NON-standard function call itoa();

Example code:

#include <stdlib.h>
#include <stdio.h>

void main( void )
{
   char buffer[20];
   int  i = 3445;
   long l = -344115L;

   _itoa( i, buffer, 10 );
   printf( "String of integer %d (radix 10): %s\n", i, buffer );
   _itoa( i, buffer, 16 );
   printf( "String of integer %d (radix 16): 0x%s\n", i, buffer );
   _itoa( i, buffer, 2  );
   printf( "String of integer %d (radix 2): %s\n", i, buffer );
}

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TriskelionCommented:
//please forgive the choppiness of this, I'm being distracted

If you want to perform math on numeric characters,
you can subtract '0' from the numeric character.

char chrSeven='7';
char chrFive='5';

int intNewVal= ((chrSeven - '0') + (chrFive - '0'));
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jkrCommented:
>>how do I get the ASCII value in C++?  (Using C++ Builder
5)

If I remember right, 'ORD()' returns the ASCII value for a character (not an integer). In C/C++ there's a direct correspondance between a char ant it's ASCII code, so by using

int nASCII = ( int) 'a';

'nASCII' will have the ASCII code for 'a'  -not keeping in mind the different codepages.

To use a library function that takes care of that, use '__toascii()':

#include <ctype.h>

int nASCII = __toascii('a');

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nietodCommented:
In C and C++, unlike in Pascal and Delphi, the character type is an integral type.  That is, it stores an integer value.  So in C/C++ you can do things like

char x = 1; // stores an interal 1.
char y = 'A'; // Stores an integral 65.
char z = x + y; // stores 66, or 'B'
char i = x*y; // stores 60, 'A'
char j = z*x; // tries to store 4290, this might overflow a char though.

So there is no need to convert an character value to an integral type, it already is an integral type.  That is why you can do.

int i = 100;
char c = 'a';

i = c;
c = i;
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