Unsigned values in C

Question

I have the following code    include  lt stdio h gt   int main         unsigned int a   -1      int b   -1      printf   x n   a       printf   x n   b        printf   d n   a       printf   d n   b        printf   u n   a       printf   u n   b       return 0      The output is   ffffffff ffffffff -1 -1 4294967295 4294967295   I can see that a value is interpreted as signed or unsigned according to the value passed to printf function  In both cases  the bytes are the same  ffffffff   Then  what is the unsigned word for

User · Answer

When you initialize unsigned int a to -1  it means that you are storing the 2 s complement of -1 into the memory of a  Which is nothing but 0xffffffff or 4294967295   Hence when you print it using  x or  u format specifier you get that output    By specifying signedness of a variable to decide on the minimum and maximum limit of value that can be stored    Like with unsigned int  the range is from 0 to 4 294 967 295 and int  the range is from -2 147 483 648 to 2 147 483 647  For more info on signedness refer this

User · Answer

In the hexadecimal it can t get a negative value  So it shows it like ffffffff   The advantage to using the unsigned version  when you know the values contained will be non-negative  is that sometimes the computer will spot errors for you  the program will  crash  when a negative value is assigned to the variable

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Assign a int -1 to an unsigned   As -1 does not fit in the range  0   UINT MAX   multiples of UINT MAX 1 are added until the answer is in range   Evidently UINT MAX is pow 2 32 -1 or 429496725 on OP s machine so a has the value of 4294967295       unsigned int a   -1    The   x     u  specifier expects a matching unsigned   Since these do not match   If a conversion specification is invalid  the behavior is undefined    If any argument is not the correct type for the corresponding conversion specification  the behavior is undefined   C11   7 21 6 1 9  The printf specifier does not change b       printf   x n   b       UB     printf   u n   b       UB   The   d  specifier expects a matching int   Since these do not match  more UB       printf   d n   a       UB   Given undefined behavior  the conclusions are not supported        both cases  the bytes are the same  ffffffff     Even with the same bit pattern  different types may have different values  ffffffff as an unsigned has the value of 4294967295   As an int  depending signed integer encoding  it has the value of -1  -2147483647 or TBD   As a float it may be a NAN      what is unsigned word for    unsigned stores a whole number in the range  0     UINT MAX    It never has a negative value   If code needs a non-negative number  use unsigned   If code needs a counting number that may be    - or 0  use int     Update  to avoid a compiler warning about assigning a signed int to unsigned  use the below   This is an unsigned 1u being negated - which is well defined as above   The effect is the same as a -1  but conveys to the compiler direct intentions   unsigned int a   -1u

User · Answer

Having unsigned in variable declaration is more useful for the programmers themselves - don t treat the variables as negative  As you ve noticed  both -1 and 4294967295 have exact same bit representation for a 4 byte integer  It s all about how you want to treat or see them   The statement unsigned int a   -1  is converting -1 in two s complement and assigning the bit representation in a  The printf   specifier x  d and u are showing how the bit representation stored in variable a looks like in different format

[c] Unsigned values in C

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