Jtdylktuy

One of my friends asked me the output of this code and I just got shocked after running this code. The output of this code is 70. Please explain why?

#include <stdio.h>

int main()



{

 int var = 0101;

 var = var+5;

 printf("%d",var);

 return 0;

}

The C Standard dictates that a numeric constant beginning with 0 is an octal constant (i.e. base-8) in § 6.4.4.1 (Integer constants).

The value 101 in base 8 is 65 in base 10, so adding 5 to it (obviously) produces 70 in base 10.

Try changing your format specifier in printf to "%o" to observe the octal representation of var.

It is because of Integer Literals. A number with leading 0 denoted that the number is an octal number. You can also use 0b for denoting binary number, for hexadecimal number it is 0x or 0X. You don't need to write any thing for decimal. See the code bellow.

#include<stdio.h>



int main()

{

    int binary = 0b10;

    int octal=010;

    int decimal = 10;

    int hexa = 0x10;

    printf("%d %d %d %dn", octal, decimal, hexa, binary);

}

For more information visit tutorialspoint.

Not shocking at all. C11 Standard - 6.4.4.1 Integer constants(p3) provides:

"An octal constant consists of the prefix 0 optionally followed by a
sequence of the digits 0 through 7 only."

Some compilers, such as gcc, provide extension for specifying binary constants, e.g. GCC Manual - 6.64 Binary Constants using the ‘0b’ Prefix But note, this is a non-standard extension.

Combining both in your example would give:

#include <stdio.h>



int main (void) {



    int var = 0101,

        bar = 0b0101;

    var = var + 5;

    bar = bar + 5;

    printf ("var: %dnbar: %dn", var, bar);



    return 0;

}

Example Use/Output

$ ./bin/octbin

var: 70

bar: 10

Inicially var is in octal numeric system, so var=0101 is equal to 001000001 in binary system or equal to 65 in decimal system.

for example in this code you can show 65 as the var inicial value.

#include <stdio.h>

int main()



{

  int var = 0101;

  printf("initial value. var=%on",var);

  var = var+5;



  printf("result of var+5. var=%dn",var);

  printf("%dn",var);

  return 0;

}

You'll get this output:

initial value. var=65

result of var+5. var=70

70