Issue
I am trying to convert some C++ code to C for my compiler that can't run with C++ code. I'd like to create the template below to C. This template converts the decimal integer to hexadecimal, and adds 0 in front of value if the size of the hexadecimal string is smaller than (sizeof(T)*2)
. Data type T
can be unsigned char
, char
, short
, unsigned short
, int
, unsigned int
, long long
, and unsigned long long
.
template< typename T > std::string hexify(T i)
{
std::stringbuf buf;
std::ostream os(&buf);
os << std::setfill('0') << std::setw(sizeof(T) * 2)
<< std::hex << i;
std::cout<<"sizeof(T) * 2 = "<<sizeof(T) * 2<<" buf.str() = "<<buf.str()<<" buf.str.c_str() = "<<buf.str().c_str()<<std::endl;
return buf.str().c_str();
}
Thank you for tour help.
Edit 1: I have tried to use the declaration
char * hexify (void data, size_t data_size)
but when I call with the int value int_value
:
char * result = hexify(int_value, sizeof(int))
it doesn't work because of:
noncompetitive type (void and int).
So in this case, do I have to use a macro? I haven't tried with macro because it's complicated.
Solution
C does not have templates. One solution is to pass the maximum width integer supported (uintmax_t
, in Value
below) and the size of the original integer (in Size
). One routine can use the size to determine the number of digits to print. Another complication is C does not provide C++’s std::string
with is automatic memory management. A typical way to handle this in C is for the called function to allocate a buffer and return it to the caller, who is responsible for freeing it when done.
The code below shows a hexify
function that does this, and it also shows a Hexify
macro that takes a single parameter and passes both its size and its value to the hexify
function.
Note that, in C, character constants such as 'A'
have type int
, not char
, so some care is needed in providing the desired size. The code below includes an example for that.
#include <inttypes.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
char *hexify(size_t Size, uintmax_t Value)
{
// Allocate space for "0x", 2*Size digits, and a null character.
size_t BufferSize = 2 + 2*Size + 1;
char *Buffer = malloc(BufferSize);
// Ensure a buffer was allocated.
if (!Buffer)
{
fprintf(stderr,
"Error, unable to allocate buffer of %zu bytes in %s.\n",
BufferSize, __func__);
exit(EXIT_FAILURE);
}
// Format the value as "0x" followed by 2*Size hexadecimal digits.
snprintf(Buffer, BufferSize, "0x%0*" PRIxMAX, (int) (2*Size), Value);
return Buffer;
}
/* Provide a macro that passes both the size and the value of its parameter
to the hexify function.
*/
#define Hexify(x) (hexify(sizeof (x), (x)))
int main(void)
{
char *Buffer;
/* Show two examples of using the hexify function with different integer
types. (The examples assume ASCII.)
*/
char x = 'A';
Buffer = hexify(sizeof x, x);
printf("Character '%c' = %s.\n", x, Buffer); // Prints "0x41".
free(Buffer);
int i = 123;
Buffer = hexify(sizeof i, i);
printf("Integer %d = %s.\n", i, Buffer); // Prints "0x00007b".
free(Buffer);
/* Show examples of using the Hexify macro, demonstrating that 'A' is an
int value, not a char value, so it would need to be cast if a char is
desired.
*/
Buffer = Hexify('A');
printf("Character '%c' = %s.\n", 'A', Buffer); // Prints "0x00000041".
free(Buffer);
Buffer = Hexify((char) 'A');
printf("Character '%c' = %s.\n", 'A', Buffer); // Prints "0x41".
free(Buffer);
}
Answered By - Eric Postpischil Answer Checked By - Marilyn (PHPFixing Volunteer)
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