Unable to convert character array elements as integer elements in C

Question

I am having character array with integer elements for example data[4]={"1234"}. I want to convert it into integer elements . There could be most simple way i.e.subtract 48 from each array element to get corresponding integer number. Here ,I am using atoi() function but not getting expected result.Current output: 1234 , 234 , 34 , 4

Expected output: 1 , 2 , 3 , 4 How should I get it?

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

int main(void)
{
  int num0;
  int num1;
  int num2;
  int num3;
  char data[6]={"1234"};

  num0 =atoi(&data[0]);
  printf("num0 =%d\n",num0);

  num1 =atoi(&data[1]);
  printf("num1 =%d\n",num1);

  num2 =atoi(&data[2]);
  printf("num2 =%d\n",num2);

  num3 =atoi(&data[3]);
  printf("num3 =%d\n",num3);
  return 0;
}

regarding; data[4]={"1234"} This is wrong. That literal "1234" takes 5 characters (remember that trailing NUL byte) — user3629249
– user3629249, Commented Aug 5, 2020 at 19:56
the posted code does not cleanly compile! running the posted code through the compiler results in: untitled1.c:12:7: warning: implicit declaration of function ‘atoi’ [-Wimplicit-function-declaration] This is because the posted code is missing the statement: #include <stdlib.h> — user3629249
– user3629249, Commented Aug 5, 2020 at 19:59
OT: when initializing an array via a literal, it is (usually) best to let the compiler determine the size of the array. Suggest: char data[] = { "1234" }; — user3629249
– user3629249, Commented Aug 5, 2020 at 20:04

Roberto Caboni · Accepted Answer · 2020-08-05 23:08:22Z

10

atoi uses all the string to do the conversion, not only the first character. Then

&data[0] is databeing "1234" => atoi(&data[0])returns 1234
&data[1] is data+1being "234" => atoi(&data[1])returns 234
&data[2] is data+2being "34" => atoi(&data[2])returns 34
&data[3] is data+3being "4" => atoi(&data[3])returns 4

you want to do something like data[i] - '0' where i is 0 .. 3

#include <stdio.h>

int main(void)
{
  char data[] = {"1234"}; /* or const char * data = "1234" */
  int i;

  for (i = 0; i != 4; i += 1)
    printf("num%d = %d\n", i+1, data[i] - '0');

  return 0;
}

Compilation and execution :

pi@raspberrypi:/tmp $ gcc -Wall c.c
pi@raspberrypi:/tmp $ ./a.out
num1 = 1
num2 = 2
num3 = 3
num4 = 4
pi@raspberrypi:/tmp $

Out of that I encourage you to stop to use atoi because atoi("aze") returns 0 without signaling an error. Instead you can use strtol or scanf with the format "%d" checking scanf returns 1.

edited Aug 5, 2020 at 23:08

Roberto Caboni

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answered Aug 5, 2020 at 18:49

bruno

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