The real difference between int and unsigned int

Question

int   The 32-bit int data type can hold integer values in the range of  -2 147 483 648 to 2 147 483 647  You may also refer to this data type  as signed int or signed   unsigned int     The 32-bit unsigned int data  type can hold integer values in the range of 0 to 4 294 967 295  You  may also refer to this data type simply as unsigned   Ok  but  in practice   int x   0xFFFFFFFF  unsigned int y   0xFFFFFFFF  printf   d   d   u   u   x  y  x  y      -1  -1  4294967295  4294967295   no difference  O o  I m a bit confused

User · Answer

The internal representation of int and unsigned int is the same   Therefore  when you pass the same format string to printf it will be printed as the same   However  there are differences when you compare them  Consider    int x   0x7FFFFFFF  int y   0xFFFFFFFF  x  lt  y    false x  gt  y    true  unsigned int  x  lt   unsigned int y     true  unsigned int  x  gt   unsigned int y     false   This can be also a caveat  because when comparing signed and unsigned integer one of them will be implicitly casted to match the types

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The binary representation is the key  An Example  Unsigned int in HEX    0XFFFFFFF   translates to   1111 1111 1111 1111 1111 1111 1111 1111    Which represents  4 294 967 295 in a base-ten positive number  But we also need a way to represent negative numbers  So the brains decided on twos complement   In short  they took the leftmost bit and decided that when it is a 1  followed by at least one other bit set to one  the number will be negative  And the leftmost bit is set to 0 the number is positive  Now let s look at what happens   0000 0000 0000 0000 0000 0000 0000 0011   3   Adding to the number we finally reach   0111 1111 1111 1111 1111 1111 1111 1111   2 147 483 645   the highest positive number with a signed integer   Let s add 1 more bit  binary addition carries the overflow to the left  in this case  all bits are set to one  so we land on the leftmost bit    1111 1111 1111 1111 1111 1111 1111 1111   -1   So I guess in short we could say the difference is the one allows for negative numbers the other does not  Because of the sign bit or leftmost bit or most significant bit

User · Answer

the type just tells you what the bit pattern is supposed to represent  the bits are only what you make of them  the same sequences can be interpreted in different ways

User · Answer

There is no difference between the two in how they are stored in memory and registers  there is no signed and unsigned version of int registers there is no signed info stored with the int  the difference only becomes relevant when you perform maths operations  there are signed and unsigned version of the maths ops built into the CPU and the signedness tell the compiler which version to use

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The problem is that you invoked Undefined Behaviour     When you invoke UB anything can happen   The assignments are ok  there is an implicit conversion in the first line  int x   0xFFFFFFFF  unsigned int y   0xFFFFFFFF    However  the call to printf  is not ok  printf   d   d   u   u   x  y  x  y     It is UB to mismatch the   specifier and the type of the argument  In your case you specify 2 ints and 2 unsigned ints in this order by provide 1 int  1 unsigned int  1 int  and 1 unsigned int     Don t do UB

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The printf function interprets the value that you pass it according to the format specifier in a matching position  If you tell printf that you pass an int  but pass unsigned instead  printf would re-interpret one as the other  and print the results that you see

User · Answer

He is asking about the real difference  When you are talking about undefined behavior you are on the level of guarantee provided by language specification - it s far from reality  To understand the real difference please check this snippet  of course this is UB but it s perfectly defined on your favorite compiler     include  lt stdio h gt   int main         int i1    0      int i2   i1  gt  gt  1      unsigned u1    0      unsigned u2   u1  gt  gt  1      printf  int            X - gt   X n   i1  i2       printf  unsigned int   X - gt   X n   u1  u2

User · Answer

Yes  because in your case they use the same representation   The bit pattern 0xFFFFFFFF happens to look like -1 when interpreted as a 32b signed integer and as 4294967295 when interpreted as a 32b unsigned integer    It s the same as char c   65  If you interpret it as a signed integer  it s 65  If you interpret it as a character it s a     As R and pmg point out  technically it s undefined behavior to pass arguments that don t match the format specifiers  So the program could do anything  from printing random values to crashing  to printing the  right  thing  etc    The standard points it out in 7 19 6 1-9     If a conversion speci cation is invalid  the behavior is unde ned  If   any argument is not the correct type for the corresponding conversion   speci cation  the behavior is unde ned

User · Answer

Hehe  You have an implicit cast here  because you re telling printf what type to expect   Try this on for size instead   unsigned int x   0xFFFFFFFF  int y   0xFFFFFFFF   if  x  lt  0      printf  one n    else     printf  two n    if  y  lt  0      printf  three n    else     printf  four n

[c] The real difference between "int" and "unsigned int"

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