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get unsigned long long addition carry

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get unsigned long long addition carry

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8

I want to get the carry bit of adding two unsigned 64-bit integers in c.
I can use x86-64 asm if needed.
code:

#include <stdio.h>

typedef unsigned long long llu;

int main(void)
 llu a = -1, b = -1;
 int carry = /*carry of a+b*/;
 llu res = a+b;
 printf("a+b = %llu (because addition overflowed), carry bit = %dn", res, carry);
 return 0;

c x86-64 addition integer-overflow extended-precision

share|improve this question

edited 53 mins ago

Peter Cordes

138k19211354

asked 4 hours ago

neo5003neo5003

441

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Can you use GCC or Clang built-ins? carry = __builtin_add_overflow(a, b, &res) stores the low bits of the result in res and sets carry to if overflow occurred. (The function actually returns a bool that is true or false, so assigning it to carry will produce 1 or 0.)
  
  – Eric Postpischil
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that msvc (if that's what you are using) also has a builtin for efficiently handling overflow (_addcarry_u64).
  
  – David Wohlferd
  1 hour ago
  

  
  
  
  

add a comment | 

8

I want to get the carry bit of adding two unsigned 64-bit integers in c.
I can use x86-64 asm if needed.
code:

#include <stdio.h>

typedef unsigned long long llu;

int main(void)
 llu a = -1, b = -1;
 int carry = /*carry of a+b*/;
 llu res = a+b;
 printf("a+b = %llu (because addition overflowed), carry bit = %dn", res, carry);
 return 0;

c x86-64 addition integer-overflow extended-precision

share|improve this question

edited 53 mins ago

Peter Cordes

138k19211354

asked 4 hours ago

neo5003neo5003

441

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Can you use GCC or Clang built-ins? carry = __builtin_add_overflow(a, b, &res) stores the low bits of the result in res and sets carry to if overflow occurred. (The function actually returns a bool that is true or false, so assigning it to carry will produce 1 or 0.)
  
  – Eric Postpischil
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that msvc (if that's what you are using) also has a builtin for efficiently handling overflow (_addcarry_u64).
  
  – David Wohlferd
  1 hour ago
  

  
  
  
  

add a comment | 

I want to get the carry bit of adding two unsigned 64-bit integers in c.
I can use x86-64 asm if needed.
code:

#include <stdio.h>

typedef unsigned long long llu;

int main(void)
 llu a = -1, b = -1;
 int carry = /*carry of a+b*/;
 llu res = a+b;
 printf("a+b = %llu (because addition overflowed), carry bit = %dn", res, carry);
 return 0;

c x86-64 addition integer-overflow extended-precision

share|improve this question

edited 53 mins ago

Peter Cordes

138k19211354

asked 4 hours ago

neo5003neo5003

441

#include <stdio.h>

typedef unsigned long long llu;

int main(void)
 llu a = -1, b = -1;
 int carry = /*carry of a+b*/;
 llu res = a+b;
 printf("a+b = %llu (because addition overflowed), carry bit = %dn", res, carry);
 return 0;

share|improve this question

edited 53 mins ago

Peter Cordes

138k19211354

asked 4 hours ago

neo5003neo5003

441

share|improve this question

edited 53 mins ago

Peter Cordes

138k19211354

asked 4 hours ago

neo5003neo5003

441

edited 53 mins ago

Peter Cordes

138k19211354

edited 53 mins ago

Peter Cordes

138k19211354

asked 4 hours ago

neo5003neo5003

441

asked 4 hours ago

neo5003neo5003

441

2 Answers
2

active

oldest

votes

5

Carry can be only 0 or 1. 1 if there was a wrapping-around and 0 otherwise.
The wrapping-around is happening in case a + b > ULONG_LONG_MAX is true . Note, this is in mathematical terms, not in terms of C, as if a + b is actually overflowing, then this will not work. Instead you want to rearrange it to be a > ULONG_LONG_MAX - b. So the value of carry will be:

carry = a > ULONG_LONG_MAX - b ? 1 : 0;

or any preferred style equivalent.

• Don't forget to include limits.h.

share|improve this answer

answered 4 hours ago

Eugene Sh.Eugene Sh.

12.6k22342

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The ? 1 : 0 is superfluous.
  
  – jxh
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jxh This is depending on the style dictated by the house rules.
  
  – Eugene Sh.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. What do you mean?
  
  – Broman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Broman By what? Some style guides tell to explicitly state the integer values resulting from conditional. Some may even require to have an if/else construct here. You are free to pick one.
  
  – Eugene Sh.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. I get it. Thanks.
  
  – Broman
  4 hours ago
  

  
  
  
  

add a comment | 

5

As @EugeneSh. observes, the carry is either 0 or 1. Moreover, given that a and b both have the same unsigned type, their sum is well defined even if the arithmetic result exceeds the range of their type. Moreover, the (C) result of the sum will be less than both a and b when overflow occurs, and greater otherwise, so we can use the fact that C relational operations evaluate to either 0 or 1 to express the carry bit as

carry = (a + b) < a;

That does not require any headers, nor does it depend on a specific upper bound, or even on a and b having the same type. As long as both have unsigned types, it reports correctly on whether the sum overflows the wider of their types or unsigned int (whichever is wider), which is the same as their sum setting the carry bit. As a bonus, it is expressed in terms of the sum itself, which I think makes it clear what's being tested.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

John BollingerJohn Bollinger

87.6k74381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks /*complete thanks message*/
  
  – neo5003
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @neo5003: you can mark an answer as accepted using the checkmark under the vote arrows.
  
  – Peter Cordes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Warning: extending this to carry_out = (a + b + carry_in) < a doesn't work, because for example b = 0xFFFF... and carry_in = 1 has already produced a carry that you won't detect with a + 0 < a being false. But yes, for add with carry-out but not carry-in, this works great. And modern compilers often can optimize c+d + carry into an adc instruction. You only run into big problems getting compilers to emit a chain of add/adc/adc/adc where you need to use the carry-out from an adc, not add.
  
  – Peter Cordes
  56 mins ago
  
  

  
  
  
  

add a comment | 

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5

Carry can be only 0 or 1. 1 if there was a wrapping-around and 0 otherwise.
The wrapping-around is happening in case a + b > ULONG_LONG_MAX is true . Note, this is in mathematical terms, not in terms of C, as if a + b is actually overflowing, then this will not work. Instead you want to rearrange it to be a > ULONG_LONG_MAX - b. So the value of carry will be:

carry = a > ULONG_LONG_MAX - b ? 1 : 0;

or any preferred style equivalent.

• Don't forget to include limits.h.

share|improve this answer

answered 4 hours ago

Eugene Sh.Eugene Sh.

12.6k22342

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The ? 1 : 0 is superfluous.
  
  – jxh
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jxh This is depending on the style dictated by the house rules.
  
  – Eugene Sh.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. What do you mean?
  
  – Broman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Broman By what? Some style guides tell to explicitly state the integer values resulting from conditional. Some may even require to have an if/else construct here. You are free to pick one.
  
  – Eugene Sh.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. I get it. Thanks.
  
  – Broman
  4 hours ago
  

  
  
  
  

add a comment | 

5

Carry can be only 0 or 1. 1 if there was a wrapping-around and 0 otherwise.
The wrapping-around is happening in case a + b > ULONG_LONG_MAX is true . Note, this is in mathematical terms, not in terms of C, as if a + b is actually overflowing, then this will not work. Instead you want to rearrange it to be a > ULONG_LONG_MAX - b. So the value of carry will be:

carry = a > ULONG_LONG_MAX - b ? 1 : 0;

or any preferred style equivalent.

• Don't forget to include limits.h.

share|improve this answer

answered 4 hours ago

Eugene Sh.Eugene Sh.

12.6k22342

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The ? 1 : 0 is superfluous.
  
  – jxh
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jxh This is depending on the style dictated by the house rules.
  
  – Eugene Sh.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. What do you mean?
  
  – Broman
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Broman By what? Some style guides tell to explicitly state the integer values resulting from conditional. Some may even require to have an if/else construct here. You are free to pick one.
  
  – Eugene Sh.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. I get it. Thanks.
  
  – Broman
  4 hours ago
  

  
  
  
  

add a comment | 

Carry can be only 0 or 1. 1 if there was a wrapping-around and 0 otherwise.
The wrapping-around is happening in case a + b > ULONG_LONG_MAX is true . Note, this is in mathematical terms, not in terms of C, as if a + b is actually overflowing, then this will not work. Instead you want to rearrange it to be a > ULONG_LONG_MAX - b. So the value of carry will be:

carry = a > ULONG_LONG_MAX - b ? 1 : 0;

or any preferred style equivalent.

• Don't forget to include limits.h.

share|improve this answer

answered 4 hours ago

Eugene Sh.Eugene Sh.

12.6k22342

carry = a > ULONG_LONG_MAX - b ? 1 : 0;

share|improve this answer

answered 4 hours ago

Eugene Sh.Eugene Sh.

12.6k22342

answered 4 hours ago

Eugene Sh.Eugene Sh.

12.6k22342

answered 4 hours ago

Eugene Sh.Eugene Sh.

12.6k22342

5

As @EugeneSh. observes, the carry is either 0 or 1. Moreover, given that a and b both have the same unsigned type, their sum is well defined even if the arithmetic result exceeds the range of their type. Moreover, the (C) result of the sum will be less than both a and b when overflow occurs, and greater otherwise, so we can use the fact that C relational operations evaluate to either 0 or 1 to express the carry bit as

carry = (a + b) < a;

That does not require any headers, nor does it depend on a specific upper bound, or even on a and b having the same type. As long as both have unsigned types, it reports correctly on whether the sum overflows the wider of their types or unsigned int (whichever is wider), which is the same as their sum setting the carry bit. As a bonus, it is expressed in terms of the sum itself, which I think makes it clear what's being tested.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

John BollingerJohn Bollinger

87.6k74381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks /*complete thanks message*/
  
  – neo5003
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @neo5003: you can mark an answer as accepted using the checkmark under the vote arrows.
  
  – Peter Cordes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Warning: extending this to carry_out = (a + b + carry_in) < a doesn't work, because for example b = 0xFFFF... and carry_in = 1 has already produced a carry that you won't detect with a + 0 < a being false. But yes, for add with carry-out but not carry-in, this works great. And modern compilers often can optimize c+d + carry into an adc instruction. You only run into big problems getting compilers to emit a chain of add/adc/adc/adc where you need to use the carry-out from an adc, not add.
  
  – Peter Cordes
  56 mins ago
  
  

  
  
  
  

add a comment | 

5

As @EugeneSh. observes, the carry is either 0 or 1. Moreover, given that a and b both have the same unsigned type, their sum is well defined even if the arithmetic result exceeds the range of their type. Moreover, the (C) result of the sum will be less than both a and b when overflow occurs, and greater otherwise, so we can use the fact that C relational operations evaluate to either 0 or 1 to express the carry bit as

carry = (a + b) < a;

That does not require any headers, nor does it depend on a specific upper bound, or even on a and b having the same type. As long as both have unsigned types, it reports correctly on whether the sum overflows the wider of their types or unsigned int (whichever is wider), which is the same as their sum setting the carry bit. As a bonus, it is expressed in terms of the sum itself, which I think makes it clear what's being tested.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

John BollingerJohn Bollinger

87.6k74381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks /*complete thanks message*/
  
  – neo5003
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @neo5003: you can mark an answer as accepted using the checkmark under the vote arrows.
  
  – Peter Cordes
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Warning: extending this to carry_out = (a + b + carry_in) < a doesn't work, because for example b = 0xFFFF... and carry_in = 1 has already produced a carry that you won't detect with a + 0 < a being false. But yes, for add with carry-out but not carry-in, this works great. And modern compilers often can optimize c+d + carry into an adc instruction. You only run into big problems getting compilers to emit a chain of add/adc/adc/adc where you need to use the carry-out from an adc, not add.
  
  – Peter Cordes
  56 mins ago
  
  

  
  
  
  

add a comment | 

As @EugeneSh. observes, the carry is either 0 or 1. Moreover, given that a and b both have the same unsigned type, their sum is well defined even if the arithmetic result exceeds the range of their type. Moreover, the (C) result of the sum will be less than both a and b when overflow occurs, and greater otherwise, so we can use the fact that C relational operations evaluate to either 0 or 1 to express the carry bit as

carry = (a + b) < a;

That does not require any headers, nor does it depend on a specific upper bound, or even on a and b having the same type. As long as both have unsigned types, it reports correctly on whether the sum overflows the wider of their types or unsigned int (whichever is wider), which is the same as their sum setting the carry bit. As a bonus, it is expressed in terms of the sum itself, which I think makes it clear what's being tested.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

John BollingerJohn Bollinger

87.6k74381

carry = (a + b) < a;

share|improve this answer

edited 4 hours ago

answered 4 hours ago

John BollingerJohn Bollinger

87.6k74381

answered 4 hours ago

John BollingerJohn Bollinger

87.6k74381

answered 4 hours ago

John BollingerJohn Bollinger

87.6k74381