Help needed in conversion

Hello,

Its been a while here now. But I had lot of help earlier.

Im facing one issue in conversion:
In the current code,

Code:

---

    long Xmin = -228725;
    long Ymax = 10023525;
    unsigned long uiRes = 50;
    long iXOffset = Xmin % uiRes;

---

iXOffset is coming as 21.

But if i change to

Code:

---

    long Xmin = -228725;
    long Ymax = 10023525;
    int uiRes = 50;
    long iXOffset = Xmin % uiRes;

---

Or
long iXOffset = Xmin % 50;

it is coming correctly as 25


Why is the current declaration of "unsigned long " causing the issue ? Please help urgently

thankyou very much
~p

To quote C99

Quote:

---

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.

Code:

---

$ 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

Thankyou so much for help. (with the details and examples ). Really helped me a lot. and its not easy to convince legacy people without proper documentation, as Im relatively new and this is legacy code, blowing up at negetive values.

Using VS2022 I get the same result as pdk5:

Code:

---

In foo: iXOffset=21
Check: -228725 == -228725
In bar: iXOffset=-25
Check: -228725 == -228725

---

PS. Note that when signed and unsigned are mixed, if possible signed is 'promoted' to unsigned. Thus signed -228725 if 'promoted' to unsigned becomes 4294738571 and 4294738571 % 50 is 21. When the divisor and dividend are both of the same sign type then no promotion is applied and mod is Xmin - (Xmin / uiRes) * uiRes) which gives 21 for 4294738571 and -25 for -228725. Note that if uiRes is -50 (ie signed), then the mod result is still -25 (you can't have an unsigned value of -50!).

Don't mix signed and unsigned - it will bite at some point and bite hard.

Thankyou very much kaud, again :) !!! .. Again great explanation for what is causing the new value..

Try this simple program:

Code:

---

#include <iostream>

int main() {
    const auto x {-4 + 2u};

    std::cout << x << '\n';
}

---

What is the type of x and what is displayed?

https://godbolt.org/z/9c14YEfPv