Help needed in conversion

[salem_c]

To quote C99

4 The result of the binary * operator is the product of the operands.
5 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. In both operations, if the value of
the second operand is zero, the behavior is undefined.
6 When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded.78) If the quotient a/b is representable, the expression
(a/b)*b + a%b shall equal a.

Note the underlined word.

You're also mixing signed and unsigned arithmetic, which is also poorly defined in C.

> it is coming correctly as 25
Weird, I get different answers.

$ cat foo.c
#include <stdio.h>

void foo ( ) {
    long Xmin = -228725;
    unsigned long uiRes = 50;
    long iXOffset = Xmin % uiRes;
    printf("In foo: iXOffset=%ld\n", iXOffset);
    printf("Check: %ld == %ld\n", Xmin, (Xmin/uiRes)*uiRes + Xmin%uiRes );
}

void bar ( ) {
    long Xmin = -228725;
    int uiRes = 50;
    long iXOffset = Xmin % uiRes;
    printf("In bar: iXOffset=%ld\n", iXOffset);
    printf("Check: %ld == %ld\n", Xmin, (Xmin/uiRes)*uiRes + Xmin%uiRes );
}

int main ( ) {
    foo();
    bar();
}

$ ./a.out 
In foo: iXOffset=41
Check: -228725 == -228725
In bar: iXOffset=-25
Check: -228725 == -228725

Both are correct, because the identity "(a/b)*b + a%b shall equal a." holds.

If you're really looking for modulo, then % is the wrong thing to be using if negative values are involved.
https://stackoverflow.com/questions/...gative-numbers
https://en.wikipedia.org/wiki/Modulo...ming_languages