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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?

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 <stdint.h>

Converting unsigned to signed integers, Learn how to find and fix bugs when converting unsigned integers to In C++ there are two types of integer variables known as signed and unsigned. a more modern C++11's brace-init syntax “int i{u};” the MSVC compiler  What is a difference between unsigned int and signed int in C? The signed and unsigned integer type has the same storage (according to the standard at least 16 bits) and alignment but still, there is a lot of difference them, in bellows lines, I am describing some difference between the signed and unsigned integer.

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*)&x;

printf("%d\n", y);

A tutorial on signed and unsigned integers « Stack Overflow, How do I convert a signed integer to an unsigned integer? This is in int a = 6; unsigned int b; int c; b = (unsigned int)a; c = (int)b;. Actually in  Let’s consider the elementary issue first. To convert a signed integer to an unsigned integer, or to convert an unsigned integer to a signed integer you need only use a cast. For example: int a = 6; unsigned int b; int c; b = (unsigned int)a; c = (int)b; Actually in many cases you can dispense with the cast.

@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.

INT02-C. Understand integer conversion rules, Both the signed char sc and the unsigned char uc are subject to integer promotions in this example. Because all values of the original types can be represented as  Signed positive values (including zero) can be stored the same way as unsigned values but since one bit is reserved for the sign the highest possible value for an n-bit number becomes 2 ^ n-1 - 1. A naive way to handle the negative values is to note if the sign bit is 1, which means that the value is negative, and then interpret the rest of the

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 <stdint.h>
   int8_t ibyte;
   uint8_t ubyte;
   int16_t iword;
   //......

What will happen when I will convert a unsigned int to , Say I want to convert an unsigned int to signed int. The operating system is 32 bit. Consider, unsigned int a=2147483649; int b; int b=(int) a; You can convert an int to an unsigned int. The conversion is valid and well-defined. Since the value is negative, UINT_MAX + 1 is added to it so that the value is a valid unsigned quantity. (Technically, 2 N is added to it, where N is the number of bits used to represent the unsigned type.)

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 > 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.

Cast unsigned INT to signed INT, POS1CNT is a 16 bit unsigned INT that is incremented/decremented When an unsigned integer is converted to a signed integer of equal What you're doing is known in the C standard as "implementation-defined behavior. 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.

A closer look at signed and unsigned integers in C., What is the difference between signed and unsigned integers? what is the size of integer? how to convert signed and unsigned? What is unsigned int in C? unsigned char is essentially a one byte of memory interpreted by the computer as an integer it is from 0 to 255. An integer type is usually 4 bytes with range -2147483648 to 2147483647. Conversion usually involves assignments from one value to another. unsigned char to integer assignment is no problem, but the other way around will have over

Online Binary-Decimal Converter, This converter allows you to convert numbers from decimal format to binary format and from binary format to decimal format. It supports the Unsigned char, 8, 0, 255, Integer. Signed char Signed int, 32, -2147483648, 2147483647, Integer. 1) Add UINT_MAX + 1 to your signed number and store it in an unsigned variable. (this will convert the signed integer to an unsigned integer of same magnitude). 2) Unsigned number of same binary representation can be found by multiplying each dig

4.5, Compare this to the 1-byte signed integer range of -128 to 127. C++ will freely convert between signed and unsigned numbers, but it won't do  Convert unsigned int to short in C. ConvertDataTypes is the helpfull website for converting your data types in several programming languages.

Comments
  • You could try this: y=(int)x<<(sizeof(int)*CHAR_BIT-16)>>(sizeof(int)*CHAR_BIT-16);. Should work on most platforms.
  • 32-bit era has come decades ago. If you're still using ancient compiler for DOS or embedded systems where int has 16 bits then the result may be as you expected
  • Why would it be -7? 65529 - 65535 = -6...
  • related stackoverflow.com/questions/50605/…
  • If you can guarantee that short is a 16-bit integer, then yes, int y = (short)x; will work. Or you can use int16_t if your compiler has the <stdint.h> header.
  • @Nayefc As I said, you can check the size of your types by printing the value of sizeof(int) sizeof(short) or whichever you need
  • There is a mostly C-compliant way to do it: y = x < 32767 ? (int)x : (x > 32768 ? -(int)-x : -32768); (the only wrinkle is that -32768 is not necessarily representable as a signed int, so you have to decide how to treat that).
  • Yeah, that works too - albeit a dirtier approach. I didn't think about going all the way to branching to do it...
  • @Mysticial: It won't branch on a decent compiler.
  • Can you please explain a little more here? Thank you.