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DDR
Asked: 2020-01-30 20:33:39 +0800 CST 2020-01-30 20:33:39 +0800 CST

Operation with numerical value of a digit does not return expected result

  • 772

I have this little code

void main()
{
    int len,i;
    int rta[64];
    char numero[64];
    scanf("%s",numero);
    len = strlen(numero);
    i=0;
    for (i=0; i<len; i++){
        rta[i]=(int)numero[i]*2;
    }
    printf(rta);
}

Whose input is 113and result should be 226, but instead it gives me the following resultb

The idea is to convert the numbers to int 's and then operate on them since I need it to read it as a string initially.

Any ideas?

Thank you!

c

1 Answers

  1. Best Answer

    You have some problems with the concept of character encoding .

    To be brief: the double of a digit does not have to match the double of that encoded digit .

    You enter 113, but actually quite a different thing is stored in memory:

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

int main( void ) {
  const char input[] = "113";

  for( size_t index = 0; index < strlen( input ); ++index ) {
    printf( "[%d]", (int)( input[index] ) );
  }

  printf( "\n" );

  return 0;
}

    Departure:

    [49] [49] [51]

    If we apply your algorithm by hand :

    #include <stdio.h>

int main( void ) {
  const char input[] = { 98, 98, 102, 0 };

  printf( "%s\n", input );

  return 0;
}

    We get the following:

    bbf

    And why does only one 'b' appear?

    This is more fun, and things like the endian of the machine come into play. To summarize, let's say that, in 32bits and little-endian , the number 49 is stored as

    00110001 00000000 00000000 00000000

    You try to read it as a character string , which is interpreted by 8 - bit packets ... and stops displaying when it reaches the first 0.

    And we still have things to discuss, such as the influence of having declared rtaas a formation of int... but we have already extended too much :-)

    Well, we're done now. Your corrected code, taking into account all the previous paragraph , would be:

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

int main( ) {
  char rta[64];
  char numero[64];

  scanf( "%s", numero );

  for( unsigned i = 0; i < strlen( numero ); ++i ) {
    rta[i] = numero[i] - '0';
    rta[i] *= 2;
    rta[i] = rta[i] + '0';
  }

  printf( "%s", rta );

  return 0;
}

    As you can see, we carry out the operation in 3 steps:

    1. We decode the digit.
    2. We perform the operation.
    3. We encode the digit, so we can display it properly.

    Note that we do not take into account the possible overflow when multiplying a digit by 2... What happens if you enter 999? :-)

    • 3