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Asked: 2026-05-26T03:01:31+00:00

This is very trivial but I am not getting it. shadyabhi@archlinux /tmp $ ./a.out

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This is very trivial but I am not getting it.

shadyabhi@archlinux /tmp $ ./a.out 
2345
51  <-- **Why?**
3
shadyabhi@archlinux /tmp $ ./a.out 
abhi
98
50  <-- **Why?**
shadyabhi@archlinux /tmp $ cat main.c 
#include <stdio.h>

int main()
{
    char a[10];
    scanf("%s", a);
    printf("%d\n", a[1]);
    printf("%d\n", a[1] - '0');
    return 0;
}
shadyabhi@archlinux /tmp $
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1 Answer

  1. Editorial Team

    When a[] contains the string "2345" as you’ve entered, then the expression a[1] is the character code for the character '3'. Since you’re using ASCII as the underlying encoding (based on the codes you’re getting, though it’s not strictly mandated by the standard), the characters '0' thru '9' are represented by the decimal values 48 thru 57 (hex 0x30 thru 0x39).

    In other words, you have:

    a[0] = '2'  = 0x32 = 50
a[1] = '3'  = 0x33 = 51
a[2] = '4'  = 0x34 = 52
a[3] = '5'  = 0x35 = 53
a[4] = '\0' = 0x00 =  0

    The ASCII table is:

        | Dec Hex Chr | Dec Hex Chr | Dec Hex Chr | Dec Hex Chr | 
    |   0  00 NUL |  32  20     |  64  40  @  |  96  60  `  | 
    |   1  01 SOH |  33  21  !  |  65  41  A  |  97  61  a  | 
    |   2  02 STX |  34  22  "  |  66  42  B  |  98  62  b  | 
    |   3  03 ETX |  35  23  #  |  67  43  C  |  99  63  c  | 
    |   4  04 EOT |  36  24  $  |  68  44  D  | 100  64  d  | 
    |   5  05 ENQ |  37  25  %  |  69  45  E  | 101  65  e  | 
    |   6  06 ACK |  38  26  &  |  70  46  F  | 102  66  f  | 
    |   7  07 BEL |  39  27  '  |  71  47  G  | 103  67  g  | 
    |   8  08 BS  |  40  28  (  |  72  48  H  | 104  68  h  | 
    |   9  09 HT  |  41  29  )  |  73  49  I  | 105  69  i  | 
    |  10  0A LF  |  42  2A  *  |  74  4A  J  | 106  6A  j  | 
    |  11  0B VT  |  43  2B  +  |  75  4B  K  | 107  6B  k  | 
    |  12  0C FF  |  44  2C  ,  |  76  4C  L  | 108  6C  l  | 
    |  13  0D CR  |  45  2D  -  |  77  4D  M  | 109  6D  m  | 
    |  14  0E SO  |  46  2E  .  |  78  4E  N  | 110  6E  n  | 
    |  15  0F SI  |  47  2F  /  |  79  4F  O  | 111  6F  o  | 
    |  16  10 DLE |  48  30  0  |  80  50  P  | 112  70  p  | 
    |  17  11 DC1 |  49  31  1  |  81  51  Q  | 113  71  q  | 
    |  18  12 DC2 |  50  32  2  |  82  52  R  | 114  72  r  | 
    |  19  13 DC3 |  51  33  3  |  83  53  S  | 115  73  s  | 
    |  20  14 DC4 |  52  34  4  |  84  54  T  | 116  74  t  | 
    |  21  15 NAK |  53  35  5  |  85  55  U  | 117  75  u  | 
    |  22  16 SYN |  54  36  6  |  86  56  V  | 118  76  v  | 
    |  23  17 ETB |  55  37  7  |  87  57  W  | 119  77  w  | 
    |  24  18 CAN |  56  38  8  |  88  58  X  | 120  78  x  | 
    |  25  19 EM  |  57  39  9  |  89  59  Y  | 121  79  y  | 
    |  26  1A SUB |  58  3A  :  |  90  5A  Z  | 122  7A  z  | 
    |  27  1B ESC |  59  3B  ;  |  91  5B  [  | 123  7B  {  | 
    |  28  1C FS  |  60  3C  <  |  92  5C  \  | 124  7C  |  | 
    |  29  1D GS  |  61  3D  =  |  93  5D  ]  | 125  7D  }  | 
    |  30  1E RS  |  62  3E  >  |  94  5E  ^  | 126  7E  ~  | 
    |  31  1F US  |  63  3F  ?  |  95  5F  _  | 127  7F DEL |

    In the second case with "abhi", a[1] is equal to b which, if you examine that table above, is character code 98 decimal ( hex 0x62).

    For the output lines where you subtract '0', that’s actually subtracting the character code for '0', which is decimal 48 or 0x30 hex.

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