What's the difference between “mod” and “remainder”?

There is a difference between modulus and remainder. For example:

-21 mod 4 is 3 because -21 + 4 x 6 is 3.

But -21 divided by 4 gives -5 with a remainder of -1.

For positive values, there is no difference.

Does '%' mean either "mod" or "rem" in C?

In C, % is the remainder¹.

..., the result of the / operator is the algebraic quotient with any fractional part discarded ... (This is often called "truncation toward zero".) C11dr §6.5.5 6

The operands of the % operator shall have integer type. C11dr §6.5.5 2

The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder ... C11dr §6.5.5 5

What's the difference between “mod” and “remainder”?

C does not define "mod", such as the integer modulus function used in Euclidean division or other modulo.
"Euclidean mod" differs from C's a%b operation when a is negative.

 // a % b 7 %  3 -->  1   7 % -3 -->  1  -7 %  3 --> -1  -7 % -3 --> -1   

"Mod" or modulo as Euclidean division

 7 modulo  3 -->  1   7 modulo -3 -->  1  -7 modulo  3 -->  2  -7 modulo -3 -->  2   

Candidate modulo code:

int modulo_Euclidean(int a, int b) {  int m = a % b;  if (m < 0) {    // m += (b < 0) ? -b : b; // avoid this form: it is UB when b == INT_MIN    m = (b < 0) ? m - b : m + b;  }  return m;}

Note about floating point: double fmod(double x, double y), even though called "fmod", it is not the same as Euclidean division "mod", but similar to C integer remainder:

The fmod functions compute the floating-point remainder of x/y. C11dr §7.12.10.1 2

fmod( 7,  3) -->  1.0  fmod( 7, -3) -->  1.0  fmod(-7,  3) --> -1.0  fmod(-7, -3) --> -1.0   

Disambiguation: C also has a similar named function double modf(double value, double *iptr) which breaks the argument value into integral and fractional parts, each of which has the same type and sign as the argument. This has little to do with the "mod" discussion here except name similarity.

[Edit Dec 2020]

For those who want proper functionality in all cases, an improved modulo_Euclidean() that 1) detects mod(x,0) and 2) a good and no UB result with modulo_Euclidean2(INT_MIN, -1). Inspired by 4 different implementations of modulo with fully defined behavior.

int modulo_Euclidean2(int a, int b) {  if (b == 0) TBD_Code(); // perhaps return -1 to indicate failure?  if (b == -1) return 0; // This test needed to prevent UB of `INT_MIN % -1`.  int m = a % b;  if (m < 0) {    // m += (b < 0) ? -b : b; // avoid this form: it is UB when b == INT_MIN    m = (b < 0) ? m - b : m + b;  }  return m;}

¹ Prior to C99, C's definition of % was still the remainder from division, yet then / allowed negative quotients to round down rather than "truncation toward zero". See Why do you get different values for integer division in C89?. Thus with some pre-C99 compilation, % code can act just like the Euclidean division "mod". The above modulo_Euclidean() will work with this alternate old-school remainder too.

sign of remainder will be same as the divisible and the sign of modulus will be same as divisor.

Remainder is simply the remaining part after the arithmetic division between two integer number whereas Modulus is the sum of remainder and divisor when they are oppositely signed and remaining part after the arithmetic division when remainder and divisor both are of same sign.

Example of Remainder:

10 % 3 = 1 [here divisible is 10 which is positively signed so the result will also be positively signed]

-10 % 3 = -1 [here divisible is -10 which is negatively signed so the result will also be negatively signed]

10 % -3 = 1 [here divisible is 10 which is positively signed so the result will also be positively signed]

-10 % -3 = -1 [here divisible is -10 which is negatively signed so the result will also be negatively signed]

Example of Modulus:

5 % 3 = 2 [here divisible is 5 which is positively signed so the remainder will also be positively signed and the divisor is also positively signed. As both remainder and divisor are of same sign the result will be same as remainder]

-5 % 3 = 1 [here divisible is -5 which is negatively signed so the remainder will also be negatively signed and the divisor is positively signed. As both remainder and divisor are of opposite sign the result will be sum of remainder and divisor -2 + 3 = 1]

5 % -3 = -1 [here divisible is 5 which is positively signed so the remainder will also be positively signed and the divisor is negatively signed. As both remainder and divisor are of opposite sign the result will be sum of remainder and divisor 2 + -3 = -1]

-5 % -3 = -2 [here divisible is -5 which is negatively signed so the remainder will also be negatively signed and the divisor is also negatively signed. As both remainder and divisor are of same sign the result will be same as remainder]

I hope this will clearly distinguish between remainder and modulus.