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I want to cast a (long) integer into a char array in C. So, for example:

int a = 65;

// I would like to get: "65"
// But I am having the conversion to the corresponding ascii when I do:

printf("%c", (char) a);
// "A"

EDIT: In my example I am printing, but what I actually want is to store the char into a variable for further manipulation.

I think this should be a super simple question, but I didnt manage to find the answer, so I thought that maybe I am not posing the question correctly. So if anyone can correct the question statement I would be grateful too.

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  • maybe you need itoa? Commented Apr 10, 2020 at 22:13

4 Answers 4

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%c string formater will output a character. Use %d to display integer instead

printf("%d", (char)a);
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7 Comments

No need for (char) casting.
Indeed, I assumed he needed it for some other reason. Casting to char is unnecessary (and error-prone) just to display a.
He probably did it because he was getting a warning from the compiler that %c requires a char.
Thank you for the answer, but I need it stored in a variable for further manipulation. So I am not sure how this translates into that, since this seems more like a formatting 'trick'?
printf only displays. If you want to store the char casted value to a variable just do char b = (char) a;. Here b will store the casted value
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I think you want to do something like:

char ch1 = (char)97;  //ch1 = 'a'
char ch2 = (char)98;  // ch2 = 'b'

printf("ch1: %c, ch2: %c\n", ch1, ch2);
printf("ch1: %c, ch2: %c\n", 97, 98);
char str[3];
// store value 97 in to str[3]
sprintf(str, "%d", 97);
printf("str: %s", str);

The test:

#include<stdio.h>

int main(void)
{
    char ch1 = (char)97;  //ch1 = 'a'
    char ch2 = (char)98;  //ch2 = 'b'

    printf("ch1: %c, ch2: %c", ch1, ch2);
    printf("ch1: %c, ch2: %c", 97, 98);

    char str[3];
     // store value 97 in to str[3]
    sprintf(str, "%d", 97);
    printf("str: %s", str);
    return 0;
}
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2 Comments

I want the opposite. I would like to have "97" and "98" into variables and not "a" nor "b"
it like: ` int a = 'a'; ` ?
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Casts are generally used for converting between fixed-size built-in datatypes with different widths or semantics (ie, extending one type to a wider one, truncating a wide type to a narrower one, or telling the compiler that we now consider something "signed" or "unsigned").

String formatting is not like a cast: strings are not a fixed-size datatype and there are many, many ways of formatting the same integer as a string (most obviously, there are multiple bases in common use).

You're already using printf, so you can just use snprintf to write the string to a char array instead of to stdout.

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Basic example of storing int variable in char array. Be aware of limits the values of various variable types like char.

#include <stdio.h>
#include <limits.h>

int main(void) {
    int a = CHAR_MAX;
    unsigned char b[1];

    b[0] = (unsigned char) a;

    printf("%d\n", a);
    printf("%hhd\n", b[0]);

    return (0);
}

Output:

➜  Desktop gcc -Wall -Werror -Wextra main.c
➜  Desktop ./a.out
127
127

If you are looking for converting your int into the char array, not just casting, then you probably need own implementation of itoa() function like this:

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

char* itoa(int val, int base){

    static char buf[32] = {0};

    int i = 30;

    for(; val && i ; --i, val /= base)

        buf[i] = "0123456789abcdef"[val % base];

    return &buf[i+1];

}

int main(void) {
    int a = 127;

    printf("Value = %s\n", itoa(a, 10));

    return (0);
}

Output:

➜  Desktop gcc -Wall -Wextra -Werror file.c
➜  Desktop ./a.out
Value = 127
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