Convert unsigned int to signed int C

Question

I am trying to convert 65529 from an unsigned int to a signed int  I tried doing a cast like this   unsigned int x   65529  int y    int  x    But y is still returning 65529 when it should return -7  Why is that

User · Answer

To understand why  you need to know that the CPU represents signed numbers using the two s complement  maybe not all  but many        byte n   1    0000 0001    1     n    n   1    1111 1110   0000 0001   1111 1111   -1   And also  that the type int and unsigned int can be of different sized depending on your CPU  When doing specific stuff like this       include  lt stdint h gt     int8 t ibyte     uint8 t ubyte     int16 t iword

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Mysticial got it  A short is usually 16-bit and will illustrate the answer   int main           unsigned int x   65529      int y    int  x      printf   d n   y        unsigned short z   65529      short zz    short z      printf   d n   zz      65529 -7 Press any key to continue          A little more detail  It s all about how signed numbers are stored in memory  Do a search for twos-complement notation for more detail  but here are the basics   So let s look at 65529 decimal  It can be represented as FFF9h in hexadecimal  We can also represent that in binary as   11111111 11111001  When we declare short zz   65529   the compiler interprets 65529 as a signed value  In twos-complement notation  the top bit signifies whether a signed value is positive or negative  In this case  you can see the top bit is a 1  so it is treated as a negative number  That s why it prints out -7   For an unsigned short  we don t care about sign since it s unsigned  So when we print it out using  d  we use all 16 bits  so it s interpreted as 65529

User · Answer

I know it s an old question  but it s a good one  so how about this   unsigned short int x   65529U  short int y     short int   amp x   printf   d n   y

User · Answer

You are expecting that your int type is 16 bit wide  in which case you d indeed get a negative value   But most likely it s 32 bits wide  so a signed int can represent 65529 just fine   You can check this by printing sizeof int

User · Answer

It seems like you are expecting int and unsigned int to be a 16-bit integer  That s apparently not the case  Most likely  it s a 32-bit integer - which is large enough to avoid the wrap-around that you re expecting   Note that there is no fully C-compliant way to do this because casting between signed unsigned for values out of range is implementation-defined  But this will still work in most cases   unsigned int x   65529  int y    short  x           If short is a 16-bit integer    or alternatively   unsigned int x   65529  int y    int16 t  x         This is defined in  lt stdint h gt

User · Answer

I know this is an old question  but I think the responders may have misinterpreted it  I think what was intended was to convert a 16-digit bit sequence received as an unsigned integer  technically  an unsigned short  into a signed integer   This might happen  it recently did to me  when you need to convert something received from a network from network byte order to host byte order   In that case  use a union   unsigned short value from network  unsigned short host val   ntohs value from network      Now suppose host val is 65529  union SignedUnsigned     short          s int    unsigned short us int     SignedUnsigned su  su us int   host val  short minus seven   su s int    And now minus seven has the value -7

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Since converting unsigned values use to represent positive numbers converting it can be done by setting the most significant bit to 0  Therefore a program will not interpret that as a Two s complement value  One caveat is that this will lose information for numbers that near max of the unsigned type   template  lt typename TUnsigned  typename TSinged gt  TSinged UnsignedToSigned TUnsigned val        return val  amp    1  lt  lt    sizeof TUnsigned    8  - 1

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The representation of the values 65529u and -7 are identical for 16-bit ints  Only the interpretation of the bits is different   For larger ints and these values  you need to sign extend  one way is with logical operations  int y    int   x   0xffff0000u      assumes 16 to 32 extension  x is  gt  32767   If speed is not an issue  or divide is fast on your processor    int y     int    x   65536u     65536    The multiply shifts left 16 bits  again  assuming 16 to 32 extension   and the divide shifts right maintaining the sign

User · Answer

To answer the question posted in the comment above - try something like this   unsigned short int x   65529U  short int y    short int x   printf   d n   y     or  unsigned short int x   65529U  short int y   0   memcpy  amp y   amp x  sizeof short int   printf   d n   y

[c] Convert unsigned int to signed int C

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