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It is very easy to convert any variable into a different kind in many languages, which gave me an idea for another converting function, which should act like str() from Python.

So what I found out is that itoa() is a function that turns an int to a char * (string):

#include <stdio.h> #include <stdlib.h> int main() { int num = 200; printf("%s", itoa(num)); }

But as it turns out, itoa() doesn't actually exist in my version of C, which they claim is C99:

make_str.c:6:18: error: implicit declaration of function 'itoa' is invalid in C99 [-Werror,-Wimplicit-function-declaration] printf("%s", itoa(num)); ^ make_str.c:6:18: error: format specifies type 'char *' but the argument has type 'int' [-Werror,-Wformat] printf("%s", itoa(num)); ~~ ^~~~~~~~~ %d

So I went to make my function instead called make_str(), though I still don't have a plan about how to convert variables into strings:

char *make_str(void *var) { }

Q: What other functions can I use to change the variables into strings?

No, not floating-point values, only int.

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  • 5
    Have you tried using sprintf()? Commented Oct 1, 2020 at 9:02
  • 1
    sprintf(), and snprintf() since C99, come to mind Commented Oct 1, 2020 at 9:03
  • 1
    itoa() is not described by any version of the C Standard, or even POSIX. Commented Oct 1, 2020 at 9:06
  • @BPML If your comment about "output functions" refers to sprintf and snprintf, you are wrong. Check the difference between printf (or fprintf) and sprintf. Commented Oct 1, 2020 at 9:30
  • To the people that marked the question a duplicate: Please read both questions before you do that. You, again, linked to a completely different question. The linked question is about Linux, this question is not about Linux, this can be relevant when the OP is on a system with very little RAM and ROM where the printf()-family needs too much RAM or ROM. Commented Oct 1, 2020 at 10:44

1 Answer 1

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In C, char variables are actually int. All of char variables have a numeric value. You can use it to convert. In ASCII, number codes are like below:
48. '0'
49. '1'
50. '2'
51. '3'
52. '4'
53. '5'
54. '6'
55. '7'
56. '8'
57. '9'

For example, the int 48 means char '0'. Thus you can use this code:

#include <stdio.h> #include <math.h> int main(void) { int entry, copy, count, dividing = 1, size = 0; char IntToChar[50]; printf("Enter an integer: "); scanf("%d", & entry); copy = entry; if (copy != 0) { while (copy != 0) { ++size; /*to calculate how many number entry has*/ copy /= 10; /*for example, if you divide 17(for int, not float) to 10, the result is 1*/ } } else size = 1; /*possibility of entry to be 0*/ for (count = 0; count < (size - 1); ++count) dividing *= 10; copy = entry; /*assignment again*/ for (count = 0; count < size; ++count) { IntToChar[count] = '0' + (copy / dividing); copy %= dividing; dividing /= 10; } IntToChar[count] = '\0'; /*adding end of string*/ printf("%s", IntToChar); return 0; }

For example,entry is 913. So, size will be 3 and dividing will be 100. If we divide copy to dividing, result is 9. And if we sum up 48('0') with this 9, it gives 57('9').After that, copy%=dividing is 13 and if we divide 13 to 10, it is 1, and we sum up this 1 with 48('0') and result is 49('1') and so on.I hope could help.

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1 Comment

Why would you want to use some 48 if you can use '0' instead? Using magic numbers makes code much worse readable and relies on character encoding.

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