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#C (gcc), (削除) 84 (削除ここまで) (削除) 82 (削除ここまで) 72 bytes

c;i;f(x){s=x;for(i=c=0;i<8;x/=10)c+=(1+2*i++%4)*x;return c%10<1;}

-21 bytes from Neil

Try it online!

Developed independently of Steadybox's answer

'f' is a function that takes the barcode as an int, and returns 1 for True and 0 for False.

• f stores the last digit of x in s (s=x%10),

• Then calculates the sum in c (for(i=c=0;i<8;x/=10)c+=(1+2*i++%4)*x;)

• c is the sum, i is a counter

• for each digit including the first, add 1+2*i%4 times the digit (x%10) to the checksum and increment i (the i++ in 3-2*i++%4)

  • 1+2*i%4 is 1 when i is even and 0 when i is odd

• Then returns whether the sum is a multiple of ten, and since we added the last digit (multiplied by 1), the sum will be a multiple of ten iff the barcode is valid.

#C (gcc), (削除) 84 (削除ここまで) (削除) 82 (削除ここまで) (削除) 72 (削除ここまで)(削除) 61 (削除ここまで) 54 bytes

c;i;f(x){for(i=c=0;x;x/=10)c+=(1+2*i++%4)*x;c=c%10<1;}

-21 bytes from Neil

-7 bytes from Nahuel Fouilleul

Try it online!

Developed independently of Steadybox's answer

'f' is a function that takes the barcode as an int, and returns 1 for True and 0 for False.

• f stores the last digit of x in s (s=x%10),

• Then calculates the sum in c (for(i=c=0;x;x/=10)c+=(1+2*i++%4)*x;)

• c is the sum, i is a counter

• for each digit including the first, add 1+2*i%4 times the digit (x%10) to the checksum and increment i (the i++ in 3-2*i++%4)

  • 1+2*i%4 is 1 when i is even and 0 when i is odd

• Then returns whether the sum is a multiple of ten, and since we added the last digit (multiplied by 1), the sum will be a multiple of ten iff the barcode is valid. (uses GCC-dependent undefined behavior to omit return).

#C (gcc), (削除) 84 (削除ここまで) (削除) 82 (削除ここまで) 72 bytes

c;s;i;f(x){s=x;for(i=c=0;i<7;x/=10,c+=(3-2*i++%4)*x);return!((c+s)%10);}

-10 bytes from Neil

Try it online!

Developed independently of Steadybox's answer

'f' is a function that takes the barcode as an int, and returns 1 for True and 0 for False.

• f stores the last digit of x in s (s=x%10),

• Then calculates the sum in c (for(i=c=0;i<7;x/=10,c+=(3-2*i++%4)*(x%10));)

• c is the sum, i is a counter

• for each digit after the first, add 3-2*i%4 times the digit (x%10) to the checksum and increment i (the i++ in 3-2*i++%4)

  • 3-2*i%4 is 3 when i is even and 1 when i is odd

• Then returns whethers is equal to the next multiple of ten minus the sum (return(c+9)%10==9-s).

#C (gcc), (削除) 84 (削除ここまで) (削除) 82 (削除ここまで) 72 bytes

c;i;f(x){s=x;for(i=c=0;i<8;x/=10)c+=(1+2*i++%4)*x;return c%10<1;}

-21 bytes from Neil

Try it online!

Developed independently of Steadybox's answer

'f' is a function that takes the barcode as an int, and returns 1 for True and 0 for False.

• f stores the last digit of x in s (s=x%10),

• Then calculates the sum in c (for(i=c=0;i<8;x/=10)c+=(1+2*i++%4)*x;)

• c is the sum, i is a counter

• for each digit including the first, add 1+2*i%4 times the digit (x%10) to the checksum and increment i (the i++ in 3-2*i++%4)

  • 1+2*i%4 is 1 when i is even and 0 when i is odd

• Then returns whether the sum is a multiple of ten, and since we added the last digit (multiplied by 1), the sum will be a multiple of ten iff the barcode is valid.

#C (gcc), (削除) 84 (削除ここまで) 82 bytes

c;s;i;f(x){s=x%10;for(i=c=0;i<7;x/=10,c+=(3-2*i++%4)*(x%10));return(c+9)%10==9-s;}

Try it online!

'f' is a function that takes the barcode as an int, and returns 1 for True and 0 for False.

• f stores the last digit of x in s (s=x%10),

• Then calculates the sum in c (for(i=c=0;i<7;x/=10,c+=(3-2*i++%4)*(x%10));)

• c is the sum, i is a counter

• for each digit after the first, add 3-2*i%4 times the digit (x%10) to the checksum and increment i (the i++ in 3-2*i++%4)

  • 3-2*i%4 is 3 when i is even and 1 when i is odd

• Then returns whether s is equal to the next multiple of ten minus the sum (return(c+9)%10==9-s).

#C (gcc), (削除) 84 (削除ここまで) (削除) 82 (削除ここまで) 72 bytes

c;s;i;f(x){s=x;for(i=c=0;i<7;x/=10,c+=(3-2*i++%4)*x);return!((c+s)%10);}

-10 bytes from Neil

Try it online!

Developed independently of Steadybox's answer

'f' is a function that takes the barcode as an int, and returns 1 for True and 0 for False.

• f stores the last digit of x in s (s=x%10),

• Then calculates the sum in c (for(i=c=0;i<7;x/=10,c+=(3-2*i++%4)*(x%10));)

• c is the sum, i is a counter

• for each digit after the first, add 3-2*i%4 times the digit (x%10) to the checksum and increment i (the i++ in 3-2*i++%4)

  • 3-2*i%4 is 3 when i is even and 1 when i is odd

• Then returns whether s is equal to the next multiple of ten minus the sum (return(c+9)%10==9-s).