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0

For some legacy reasons I have code that casts a double to unsigned byte and we are seeing much difference between the two platforms. Short of doing - "don't try to stuff a signed value into unsigned value; don't try to stuff a double value in integer", is there something else that can be done?

unsigned char newR1 = -40.16;

Value of newR1 is 216 on windows (as we expected from a long time); but on ARM64 it is 0.

Disassembly on Win:

00007FF75E388818 cvttsd2si eax,mmword ptr [R] 
00007FF75E38881D mov byte ptr [newR1],al

On ARM64

00007FF6F9E800DC ldr d16,[sp,#0x38 |#0x38 ] 
00007FF6F9E800E0 fcvtzu w8,d16 
00007FF6F9E800E4 uxtb w8,w8 
00007FF6F9E800E8 strb w8,[sp,#0x43 |#0x43 ]

Will try these as well, but just wanted some other opinions

unsigned char newR1 = -40.16;
unsigned char newR2 = (int)-40.16;
unsigned char newR3 = (unsigned char)-40.16;
unsigned char newR4 = static_cast<int>(-40.16);

or may be

int i = -40.16;
unsigned char c = i;

c++ arm

share|improve this question

edited Mar 8 at 4:41

S.M.

6,30241728

asked Mar 8 at 4:25

AmitAmit

1,543717

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What value do you expect to see?
  
  – Mark Ransom
  Mar 8 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well 216; -40.16 first become -40; which then becomes 216 preserving the bit pattern representation of -40. That is what everyone except ARM64 is doing - Windows, iOS, Mac. That is the behavior (may be undefined, is it?) I have to come to expect when casting signed/unsigned.
  
  – Amit
  Mar 8 at 4:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  what compiler and version?
  
  – old_timer
  Mar 8 at 4:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  so.c:5:12: warning: overflow in conversion from ‘double’ to ‘unsigned char’ changes value from ‘-4.0159999999999997e+1’ to ‘0’ [-Woverflow]
  
  – old_timer
  Mar 8 at 4:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Visual Studio 2017 ARM64 compiler targeting Windows machines.
  
  – Amit
  Mar 8 at 4:45
  

  
  
  
  

 | 
show 2 more comments

0

For some legacy reasons I have code that casts a double to unsigned byte and we are seeing much difference between the two platforms. Short of doing - "don't try to stuff a signed value into unsigned value; don't try to stuff a double value in integer", is there something else that can be done?

unsigned char newR1 = -40.16;

Value of newR1 is 216 on windows (as we expected from a long time); but on ARM64 it is 0.

Disassembly on Win:

00007FF75E388818 cvttsd2si eax,mmword ptr [R] 
00007FF75E38881D mov byte ptr [newR1],al

On ARM64

00007FF6F9E800DC ldr d16,[sp,#0x38 |#0x38 ] 
00007FF6F9E800E0 fcvtzu w8,d16 
00007FF6F9E800E4 uxtb w8,w8 
00007FF6F9E800E8 strb w8,[sp,#0x43 |#0x43 ]

Will try these as well, but just wanted some other opinions

unsigned char newR1 = -40.16;
unsigned char newR2 = (int)-40.16;
unsigned char newR3 = (unsigned char)-40.16;
unsigned char newR4 = static_cast<int>(-40.16);

or may be

int i = -40.16;
unsigned char c = i;

c++ arm

share|improve this question

edited Mar 8 at 4:41

S.M.

6,30241728

asked Mar 8 at 4:25

AmitAmit

1,543717

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What value do you expect to see?
  
  – Mark Ransom
  Mar 8 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well 216; -40.16 first become -40; which then becomes 216 preserving the bit pattern representation of -40. That is what everyone except ARM64 is doing - Windows, iOS, Mac. That is the behavior (may be undefined, is it?) I have to come to expect when casting signed/unsigned.
  
  – Amit
  Mar 8 at 4:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  what compiler and version?
  
  – old_timer
  Mar 8 at 4:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  so.c:5:12: warning: overflow in conversion from ‘double’ to ‘unsigned char’ changes value from ‘-4.0159999999999997e+1’ to ‘0’ [-Woverflow]
  
  – old_timer
  Mar 8 at 4:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Visual Studio 2017 ARM64 compiler targeting Windows machines.
  
  – Amit
  Mar 8 at 4:45
  

  
  
  
  

 | 
show 2 more comments

For some legacy reasons I have code that casts a double to unsigned byte and we are seeing much difference between the two platforms. Short of doing - "don't try to stuff a signed value into unsigned value; don't try to stuff a double value in integer", is there something else that can be done?

unsigned char newR1 = -40.16;

Value of newR1 is 216 on windows (as we expected from a long time); but on ARM64 it is 0.

Disassembly on Win:

00007FF75E388818 cvttsd2si eax,mmword ptr [R] 
00007FF75E38881D mov byte ptr [newR1],al

On ARM64

00007FF6F9E800DC ldr d16,[sp,#0x38 |#0x38 ] 
00007FF6F9E800E0 fcvtzu w8,d16 
00007FF6F9E800E4 uxtb w8,w8 
00007FF6F9E800E8 strb w8,[sp,#0x43 |#0x43 ]

Will try these as well, but just wanted some other opinions

unsigned char newR1 = -40.16;
unsigned char newR2 = (int)-40.16;
unsigned char newR3 = (unsigned char)-40.16;
unsigned char newR4 = static_cast<int>(-40.16);

or may be

int i = -40.16;
unsigned char c = i;

c++ arm

share|improve this question

edited Mar 8 at 4:41

S.M.

6,30241728

asked Mar 8 at 4:25

AmitAmit

1,543717

unsigned char newR1 = -40.16;

00007FF75E388818 cvttsd2si eax,mmword ptr [R] 
00007FF75E38881D mov byte ptr [newR1],al

00007FF6F9E800DC ldr d16,[sp,#0x38 |#0x38 ] 
00007FF6F9E800E0 fcvtzu w8,d16 
00007FF6F9E800E4 uxtb w8,w8 
00007FF6F9E800E8 strb w8,[sp,#0x43 |#0x43 ]

unsigned char newR1 = -40.16;
unsigned char newR2 = (int)-40.16;
unsigned char newR3 = (unsigned char)-40.16;
unsigned char newR4 = static_cast<int>(-40.16);

int i = -40.16;
unsigned char c = i;

share|improve this question

edited Mar 8 at 4:41

S.M.

6,30241728

asked Mar 8 at 4:25

AmitAmit

1,543717

share|improve this question

edited Mar 8 at 4:41

S.M.

6,30241728

asked Mar 8 at 4:25

AmitAmit

1,543717

edited Mar 8 at 4:41

S.M.

6,30241728

edited Mar 8 at 4:41

S.M.

6,30241728

asked Mar 8 at 4:25

AmitAmit

1,543717

asked Mar 8 at 4:25

AmitAmit

1,543717

1 Answer
1

active

oldest

votes

4

What the C standard says (and there's similar text in the C++ one):

When a finite value of real floating type is converted to an integer
type other than _Bool, the fractional part is discarded (i.e., the
value is truncated toward zero). If the value of the integral part
cannot be represented by the integer type, the behavior is undefined.

So, getting 216 out of -40.16 with a single cast from double to unsigned char is already UB. In fact, getting any result in this case is UB. Which is why the compiler is free to produce anything and not 216 that you desire.

You may want to do two casts:

(unsigned char)(int)-40.16

Again, the first cast (to int) is still subject to the above restriction I quoted.

share|improve this answer

answered Mar 8 at 5:27

Alexey FrunzeAlexey Frunze

52.7k953129

add a comment | 

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4

What the C standard says (and there's similar text in the C++ one):

When a finite value of real floating type is converted to an integer
type other than _Bool, the fractional part is discarded (i.e., the
value is truncated toward zero). If the value of the integral part
cannot be represented by the integer type, the behavior is undefined.

So, getting 216 out of -40.16 with a single cast from double to unsigned char is already UB. In fact, getting any result in this case is UB. Which is why the compiler is free to produce anything and not 216 that you desire.

You may want to do two casts:

(unsigned char)(int)-40.16

Again, the first cast (to int) is still subject to the above restriction I quoted.

share|improve this answer

answered Mar 8 at 5:27

Alexey FrunzeAlexey Frunze

52.7k953129

add a comment | 

4

What the C standard says (and there's similar text in the C++ one):

When a finite value of real floating type is converted to an integer
type other than _Bool, the fractional part is discarded (i.e., the
value is truncated toward zero). If the value of the integral part
cannot be represented by the integer type, the behavior is undefined.

So, getting 216 out of -40.16 with a single cast from double to unsigned char is already UB. In fact, getting any result in this case is UB. Which is why the compiler is free to produce anything and not 216 that you desire.

You may want to do two casts:

(unsigned char)(int)-40.16

Again, the first cast (to int) is still subject to the above restriction I quoted.

share|improve this answer

answered Mar 8 at 5:27

Alexey FrunzeAlexey Frunze

52.7k953129

add a comment | 

What the C standard says (and there's similar text in the C++ one):

When a finite value of real floating type is converted to an integer
type other than _Bool, the fractional part is discarded (i.e., the
value is truncated toward zero). If the value of the integral part
cannot be represented by the integer type, the behavior is undefined.

So, getting 216 out of -40.16 with a single cast from double to unsigned char is already UB. In fact, getting any result in this case is UB. Which is why the compiler is free to produce anything and not 216 that you desire.

You may want to do two casts:

(unsigned char)(int)-40.16

Again, the first cast (to int) is still subject to the above restriction I quoted.

share|improve this answer

answered Mar 8 at 5:27

Alexey FrunzeAlexey Frunze

52.7k953129

(unsigned char)(int)-40.16

share|improve this answer

answered Mar 8 at 5:27

Alexey FrunzeAlexey Frunze

52.7k953129

answered Mar 8 at 5:27

Alexey FrunzeAlexey Frunze

52.7k953129

answered Mar 8 at 5:27

Alexey FrunzeAlexey Frunze

52.7k953129