Integer division in C with unsigned short

Question

I feel like the bloodiest beginner - Why does the following not work:

// declarations
unsigned short currentAddr= 0x0000;
unsigned short addr[20] = {1, 0};

// main 
addr[1] = (~currentAddr)/2+1;
printf("addr[1] wert: %hu\n", addr[1]); // equals 1, expecte 0x8000
addr[1] = ~currentAddr>>1;
printf("addr[1] wert: %hu\n", addr[1]); // equals 65535, expected 0x7FFF

In printf and also in my debugger's watchlist the value for addr[1] is not as expected. My aim is to have half the maximum of the variable, here 0x8000. Info: I am doing ~currentAddr to get the max. 0xFFFF in case short is in a different length on my embedded platform than here on my PC.

cheers, Stefan

Answer 1

What went wrong

The integer promotions are performed on the operand of the unary ~.

On many systems int is larger than short. On such systems, for unsigned short currentAddr = 0, the value of currentAddr is first promoted to int in the expression ~currentAddr. Then ~currentAddr evaluates to -1 (assuming twos-complement representation).

On some systems int and short may be the same size (though int must be at least as large as short); here currentAddr would instead be promoted to unsigned int since an int cannot hold all values of an unsigned integer type of the same size. In such a case, ~currentAddr would evaluate to UINT_MAX. For 16-bit int (short must be at least 16-bit, so here int and short would be the same size) the result of ~currentAddr would be 65,535.

The OP's system must have int larger than short. In the case of addr[1] = (~currentAddr)/2+1; this becomes addr[1] = (-1)/2+1; which evaluates to 1.

In the second case, addr[1] = ~currentAddr>>1; evaluates to addr[1] = (-1)>>1;. Here, the result of right-shifting a negative value is implementation-defined. In the present case, the result appears to be INT_MAX, which is converted to unsigned short in the assignment to addr[1], which takes the value USHRT_MAX in the conversion. This value is 65,535 on OP's system.

What to do about it

To obtain maximum and minimum values for standard integer types clearly and reliably, use the macros found in limits.h instead of attempting bit manipulations. This method will not disappoint:

#include <stdio.h>
#include <limits.h>

int main(void)
{
    unsigned short val;

    val = (USHRT_MAX / 2) + 1;
    printf("(USHRT_MAX / 2) + 1: %#hx\n", val);

    val = USHRT_MAX >> 1;
    printf("     USHRT_MAX >> 1: %#hx\n", val);

    return 0;
}

Program output:

(USHRT_MAX / 2) + 1: 0x8000
     USHRT_MAX >> 1: 0x7fff

Answer 2

The problem lies here:

addr[1] = (~currentAddr)/2+1;

You expect currentAddr to 0xFFFF, which is partially right. But, what you might have missed out is integer promotion rule, which makes it 0xFFFFFFFF which is hexadecimal representation of -1.

Now, there is simple math: (~currentAddr)/2+1 is nothing but 0x01 or 1, When you ~currentAddr>>1; do this shift, it is again becoming -1.

From

My aim is to have half the maximum of the variable, here 0x8000

If I understand you correctly, what you are trying to do is get the value which is equal to (Maximum value of Unsigned short)/2. If it is so, the proper way of doing it will be using USHRT_MAX. Of course, you'll need to include limits.h file in your source code.

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Update:

Referring to your comments to David's answer, following changes works as expected. (You have tested, I haven't)

unsigned short c; 
c = ~currentAddr; 
unsigned short c_z = sizeof (c);
unsigned short ci; 
ci = (c >> 1) + 1; 
unsigned short ci_z = sizeof (ci); 
addr[1] = ci;

Now, why this isn't promoted to integer as opposed to previous case

c = ~currentAddr; 

It is promoted, but it yields an expected result because, as chux explainned (which I couldn't have done) it is (temporarily) promoted to int during its operation, but resolved as (converted to) a unsigned short again when it is stored in memory allocated to c.

The C standard answers the question:

From the C99 standard: 6.5.16.1 Simple assignment

In simple assignment (=), the value of the right operand is converted to the type of the assignment expression and replaces the value stored in the object designated by the left operand. In your case since both the LHS and RHS are of the same type, there is no need for any conversion.

Also, it says:

The type of an assignment expression is the type the left operand would have after lvalue conversion.

The same is specified by C11 6.5.16.1/2:

In simple assignment (=), the value of the right operand is converted to the type of the assignment expression and replaces the value stored in the object designated by the left operand.

Try this yourself:

int main(void)
{
    unsigned short c; 
    unsigned short currentAddr= 0x0000;
    c = ~currentAddr;

    printf("\n0x%x", c);
    printf("\n0x%x", (~currentAddr));

    return 0;
}

This should print:

0xffff
0xffffffff

Answer 3

addr[1] = (~currentAddr)/2+1;

Let us break it down: currentAddr is an unsigned short involved in a computation so the value/type is first promoted to int or unsigned. In C this is integer promotion.

If an int can represent all values of the original type ..., the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions. C11dr §6.3.1.1 2

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When USHRT_MAX <= INT_MAX, (e.g. 16 bit short int/unsigned, 32-bit int/unsigned), code is like below. With currentAddr == 0 and typical 2's complement behavior, ~0 --> -1 and addr[1] --> 1.

int tmp = currentAddr;
addr[1] = (~tmp)/2+1;  

When USHRT_MAX > INT_MAX, (e.g. 16 bit short int/unsigned, 16-bit int/unsigned), code is like below. With currentAddr == 0 and unsigned behavior, ~0 --> 0xFFFF and addr[1] --> 0x8000.

unsigned tmp = currentAddr;
addr[1] = (~tmp)/2+1;  

---

My aim is to have half the maximum of the variable

The best way to get the maximum of an unsigned short is to use SHRT_MAX and skip the ~ code. It will work as expected regardless of unsigned short, int, unsigned range. It also better documents code intent.

#include <limits.h>
addr[1] = USHRT_MAX/2+1;

Answer 4

Because the number 2 is int and int can hold unsigned short,so,the the actual operation is addr[1] = (unsigned short)(((int)(~currentAddr)/2)+1)