bit shifting with unsigned long type produces wrong results

by

1 << 63 will be computed in int arithmetic, and your int is probably 32 bit.

Remedy this by promoting one of the arguments: 1ULL << 63 will do it.

ULL means the expression will be at least 64 bits.

More Related Contents:

  1. C – is any use of unsigned int just terrible coding practice? [closed]
  2. Comparison operation on unsigned and signed integers
  3. Arithmetic bit-shift on a signed integer
  4. Are the shift operators () arithmetic or logical in C?
  5. Is unsigned integer subtraction defined behavior?
  6. Right shifting negative numbers in C
  7. What does the C standard say about bitshifting more bits than the width of type?
  8. Bitwise operators and “endianness”
  9. Why is a negative int greater than unsigned int? [duplicate]
  10. Difference between signed / unsigned char [duplicate]
  11. How to printf “unsigned long” in C?
  12. Difference between unsigned and unsigned int in C
  13. Difference between size_t and unsigned int?
  14. Why doesn’t C have unsigned floats?
  15. Unsigned hexadecimal constant in C?
  16. What does the statement if (counter & (1
  17. c get nth byte of integer
  18. Unsigned values in C
  19. C left shift on 64 bits fail
  20. What happens when I assign a negative value to an unsigned int? [duplicate]
  21. what is the unsigned datatype?
  22. warning: left shift count >= width of type
  23. C Unsigned int providing a negative value?
  24. C reverse bits in unsigned integer
  25. Why is int rather than unsigned int used for C and C++ for loops?
  26. Why use the Bitwise-Shift operator for values in a C enum definition?
  27. Why does combining two shifts of a uint8_t produce a different result?
  28. Declaring 64-bit variables in C
  29. OpenMP time and clock() give two different results
  30. Increasing camera capture resolution in OpenCV

Leave a Comment