Cast an int to char

**tinchi** · 10-02-2014

Is it possible to get the string representation of an int. So I mean if i have:

5, I want to get '5'.

I was searching in the internet and I found this solution:

Code:

int number = 5;
char c = number + '0';

which works fine for small numbers, but if I do:

Code:

int i;
char c;
for(i =0; i < 10000; i++) {
c = i + '0';
printf("%c\n", c);
}

At some point some strange sings like %&/)/)§$%&) are appearing, But i really need this huge numbers as well .. Is there any other way to do this??

**Matticus** · 10-02-2014

Look into the function "sprintf()".

(Note: Your examples deal with single characters, not strings.)

**grumpy** · 10-02-2014

Assuming you have something like;

Code:

   int value = 10000;
   char buffer[some_big_enough_value];     /*  for value of 10000, some_big_enough_value must be 6 or more */

Try

Code:

   sprintf(buffer, "%d", value);      /*  sprintf(), like printf() is declared in <stdio.h> */
   printf("%s\n", buffer);

or

Code:

   itoa(value, buffer,10);      /*  itoa() is declared in <stdlib.h> */
   printf("%s\n", buffer);

or even (since itoa() returns its second argument)

Code:

   printf("%s\n", itoa(value, buffer,10));

**tinchi** · 10-02-2014

Undefined symbols for architecture x86_64:
"_itoa", referenced from:
_main in main.o
ld: symbol(s) not found for architecture x86_64

But sprintf does it for me. thank you really much!!!!!!!!

**tinchi** · 10-02-2014

And what if I want to do it the other direction, is an (int) enough?
Wow this is a really nice method. thank you a lot. I also managed to cast a address to a char:

Code:

    int lala = 0;
    
    printf("%p \n", &lala);
    
    char buffer[100];
    sprintf(buffer, "%p", &lala);
    printf("%s\n", buffer);

Can you give me some advice how to do these things but in the other direction? so char -> pointer, char -> int

**the_jackass** · 10-02-2014

I know it's off-topic but could u please look into this thrad: Command line argument parsing issues?

**the_jackass** · 10-02-2014

Originally Posted by tinchi

And what if I want to do it the other direction, is an (int) enough?

Nope. atoi() is modulo 2^32. Ex:

Code:

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

int main() {
 unsigned int v=atoi("9876543210");
	printf("%u\n",v);
	return 0;
}

gives 1286608618 which is 9876543210%(1<<32)
code that worx upto 64 bit (unsigned long long) can be written by you (no library function exists on that size)

greater than that you'll need arbritary precision numbers. (If you dont know what that is, dont ask and be satisfied with numbers that fit in ints)

**cas** · 10-02-2014

atoi() is modulo 2^32

atoi() is undefined if the result does not fit into an int; lots of systems will reduce it modulo 2**32, but that's not a requirement. Also, int need not be 32 bits wide (it could be smaller or larger).

code that worx upto 64 bit (unsigned long long) can be written by you (no library function exists on that size)

In C99 (which introduced (unsigned) long long to standard C), there exist the functions strtoll() and strtoull(), which convert a string to long long/unsigned long long. In addition, they have more advantages over atoi(): overflow is not undefined and is detectable; you can specify the base; and you can determine where exactly the parsing stopped, so you can tell if the entire string was converted.

**the_jackass** · 10-03-2014

Originally Posted by cas

In C99 (which introduced (unsigned) long long to standard C), there exist the functions strtoll() and strtoull(), which convert a string to long long/unsigned long long. In addition, they have more advantages over atoi(): overflow is not undefined and is detectable; you can specify the base; and you can determine where exactly the parsing stopped, so you can tell if the entire string was converted.

Cool. I'll have to look into that someday.

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