## Printing the actual negative number stored inside unsigned int

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I have a strange problem. I have a variable whose actual value is only negative(only negative integers are generated for this variable). But in the legacy code, an old colleague of mine used uint16 instead of signed int to store the values of that variable. Now if i wanted to print the actual negative value of that variable, how can i do that(what format specifier to us)? For example if actual value is -75, when i print using %d its giving me 5 digit positive value(I think its because of two's complement). I want to print it as 75 or -75.

**Can unsigned numbers stores negative values?,** As the name suggests, we can store both positive and negative numbers. Sanket Savla, Senior Consultant at Capgemini India (2017-present) as positive value in unsigned integer, but you can assign signed integer to unsigned integer. it is multiplied by 3 repeating the statement until integer is 1 and print the count. Unsigned integer range. A 1-byte unsigned integer has a range of 0 to 255. Compare this to the 1-byte signed integer range of -128 to 127. Both can store 256 different values, but signed integers use half of their range for negative numbers, whereas unsigned integers can store positive numbers that are twice as large.

If an `int`

is 32 bits on your system, then the correct format for printing a 16-bit int is `%hu`

for unsigned and `%hd`

for signed.

Examine the output of:

uint16_t n = (uint16_t)-75; printf("%d\n", (int)n); // Output: 65461 printf("%hu\n", n); // Output: 65461 printf("%hd\n", n); // Output: -75

**4.5 — Unsigned integers, and why to avoid them,** Both can store 256 different values, but signed integers use half of their range for When no negative numbers are required, unsigned integers are well-suited The number 280 is too big to fit in our 1-byte range of 0 to 255. with this. instead of printing nothing it will wrap-around and show 4294967295. The %d format treats its argument as a signed int.Use %u instead.. But a better way to get the maximum value of type unsigned int is to use the UINT_MAX macro. You'll need. #include <limits.h>

#include <inttypes.h> uint16_t foo = -75; printf("==> %" PRId16 " <==\n", foo); // type mismatch, undefined behavior

**How the negative numbers are stored in memory?,** Suppose the following fragment of code, int a = -34; Now how will this be stored in memory. So here is the complete theory. Whenever a number with minus sign � The range of char is 127 to -128. If you assign 212, ch stores -44 (212-128-128) not 212.So if you try to print a negative number as unsigned you get (MAX value of unsigned int)-abs (number) which in this case is 4294967252 So if you want to store 212 as it is in ch the only thing you can do is declare ch as

Assuming your friend has somehow correctly put the bit representation for the signed integer into the unsigned integer, the only standard compliant way to extract it back would be to use a union as-

typedef union { uint16_t input; int16_t output; } aliaser;

Now, in your code you can do -

aliaser x; x.input = foo; printf("%d", x.output);

**Exam 1 Review,** In 5 bits, what is the most negative value representable in unsigned form? x <= MAX_UNSIGNED_CHAR will stay true for any of the possible values x gets. you should use special functionality provided by SPIM to print the integer stored in� The argument is interpreted as a short int or unsigned short int (only applies to integer specifiers: i, d, o, u, x and X). 2: l. The argument is interpreted as a long int or unsigned long int for integer specifiers (i, d, o, u, x and X), and as a wide character or wide character string for specifiers c and s. 3: L

Go ahead with the union, as explained by another answer. Just wanted to say that if you are using GCC, there's a very cool feature that allows you to do that sort of "bitwise" casting without writing much code:

printf("%d", ((union {uint16_t in; int16_t out}) foo).out);

See https://gcc.gnu.org/onlinedocs/gcc-4.4.1/gcc/Cast-to-Union.html.

**Signed number representations,** In computing, signed number representations are required to encode negative numbers in For integers, the representation used in most current computing devices is two's complement, although the the bit pattern which is one greater ( in an unsigned sense) than the ones' complement of the positive value. Print/ export. C Program to Print an Integer (Entered by the User) In this example, the integer entered by the user is stored in a variable and printed on the screen. To understand this example, you should have the knowledge of the following C programming topics:

**Integer overflow,** In computer programming, an integer overflow occurs when an arithmetic operation attempts to The C11 standard states that for unsigned integers modulo wrapping is the defined large enough to contain all values that may possibly be computed and stored in them, it is always possible to avoid overflow. Print/export. The width argument is a non-negative decimal integer that controls the minimum number of characters that are output. If the number of characters in the output value is less than the specified width, blanks are added to the left or the right of the values—depending on whether the left-alignment flag ( - ) is specified—until the minimum width

**Lecture notes - Chapter 5,** EXAMPLES: 11100 this must be a negative number. to find out which, find the additive inverse! get an integer in 2's comp. representation: positive values are the same as for TRUE VALUE to be represented 3 add in the bias +8 ---- unsigned value 11 Then, do the arithmetic on integers, and print out the integer result. Therefore, a signed integer can hold both positive and negative numbers (and 0). In this lesson, we’ll focus on signed integers. We’ll discuss unsigned integers (which can only hold non-negative numbers) in the next lesson. Defining signed integers. Here is the preferred way to define the four types of signed integers:

**A tutorial on signed and unsigned integers � Stack Overflow,** In my case, after I posted these thoughts on signed versus unsigned integers, How do I convert a signed integer to an unsigned integer? if (a < 0) { res = true; /* If a is negative, it must be less than b */ } else { unsigned int c; I want to print counter increments and decrements which varies from -120 to� Unsigned integers can represent zero and positive integers, but not negative integers. The value of an unsigned integer is interpreted as " the magnitude of its underlying binary pattern ". Example 1: Suppose that n =8 and the binary pattern is 0100 0001B , the value of this unsigned integer is 1×2^0 + 1×2^6 = 65D .

##### Comments

- Cast the variable in the print call.
`printf("value = %d\n", (int)value);`

- What is with all the answers attempting to use fancy features with unions or specifier abuse? Just subtract 65536. Done. Standard, defined, simple.
- It'd help a lot if we knew details, in particular what is sizeof(int). Whether it's 2 or 4 makes quite a difference.
- You might want to add the note that this will only work if
`sizeof(int) > 2`

. There are still compilers out there using 16 bit ints. Mostly in the embedded space, e.g. the Arduino: arduino.cc/reference/en/language/variables/data-types/int - @dgnuff: If
`int`

is 16 bits, then`65536`

is a long, and, in`x-65536`

,`x`

is converted to a long, and then the resulting value is correct. - @EricPostpischil: The Standard makes no attempt to forbid implementations from behaving in obtuse fashion. Unless one is using an unusual platform, an obtuse compiler, or a compiler that's configured to squawk at actions which are defined but unusual,
`(int16_t)65461`

is going to equal -75. A bigger issue from a practical perspective would be that not all printf libraries implement everything included in the spec. - @EricPostpischil: The Standard has never attempted to specify everything necessary to make an implementation suitable for any particular purpose. Indeed, the authors explicitly note in the Rationale that an implementation could be "conforming" without being suitable for any particular purpose. When the authors of the C Standard described in the Rationale how "the majority of current implementations" would process
`unsigned short a=60000,b=62500; unsigned c=a*b;`

, did they mean that the majority of implementations explicitly documented such behavior, or that... - ...they thought that the tradition of how such actions should be handled on commonplace platforms was sufficiently strong and unambiguous that there was no reason to contemplate the notion of compilers doing anything else except when targeting unusual platforms where other behaviors might make more sense?
- @EricPostpischil: PS--while it's possible that an implementation might define
`uint16_t`

and yet define the`short int`

as a different type from`int16_t`

, such behavior would be sufficiently unusual that I'd be surprised if*any*non-contrived implementations of C99 or later have ever done so. - @EricPostpischil: My "personal opinions about what compilers ought to do" are, in this case, backed up by decades observations of what general-purpose compilers for commonplace platforms have historically,
*unanimously*, done. While there are a number of ways of interpreting an attempt to convert an out-of-range value to a small signed integer type, and they would yield different results when using ones'-complement or sign-magnitude representations, they all yield the same value when using two's-complement representation. `PRI16D`

is defined for formatting an`int16_t`

, not for reinterpreting the value of a`uint16_t`

. The C standard does not define the behavior when used in this way.`Assuming your friend has somehow correctly put the bit representation for the signed integer into the unsigned integer`

<- this is as simple as`T foo = (T)-42;`

- This is not the only standard compliant way. Simply subtracting 65536 restores the original value.