## about the largest value an int can hold in c programming?

unsigned long max value

long long max value c++

max number in c

int max

int_max

max number in c++

double>::max value c++

In my system, the size of int datatype is 4 bytes so the maximum unsigned int value it can have is 4294967296.

# include<stdio.h> int main(void) { unsigned int a = 4294967296; return 0; }

why my compiler throwing an error even though the maximum value it can hold is 4294967296? the warning is, " overflow in implicit constant conversion"

thank you!!

You do not need to remember such constants by heart or calculate them yourself. The largest `int`

value is defined in the file `limits.h`

. The file also defines the largest `short`

, `unsigned int`

, etc., as well as the smallest values for each integral data type:

#include <limits.h> ... int a = INT_MAX; short b = SHRT_MAX; unsigned int c = UINT_MAX;

**INT_MAX and INT_MIN in C/C++ and Applications,** ) is the maximum positive value for a 32-bit signed binary integer in computing. It is therefore the maximum value for variables declared as integers (e.g., as int ) in many programming languages, and the maximum possible score, money, etc. for many video games. Microsoft C also permits the declaration of sized integer variables, which are integral types of size 8-, 16-, 32- or 64-bits. For more information on sized integers in C, see Sized Integer Types. Limits on Integer Constants. Number of bits in the smallest variable that is not a bit field. Minimum value for a variable of type signed char.

`int`

is a signed value, so it goes to negative too which effectively halves that amount in the positive realm. That warning is telling you that it will overflow into a negative number.

**2,147,483,647,** What is the maximum value of unsigned int in C? As tabstop mentioned,you include the limits.h header file in your program.You can simply the print the values by using the macros such as INT_MAX to get the maximum value and INT_MIX to get the minimum value.

*even though the maximum value it can hold is 4294967296 ?*

**No**, it can't as

`a`

is default `signed`

in your case. `INT_MAX`

is `2147483647`

not `4294967296`

(it's `UINT_MAX + 1`

) So when you do

int a = 4294967296; /* it can hold max 2147483647 on 32-bit, more than this cause overflow */

it will overflow into a negative number(`INT_MIN`

to `INT_MAX`

) & also the behavior on signed integer overflow is **undefined** by the C standard.

**C Programming/limits.h,** What is the data type that occupies the most storage in C language? The following table gives you the common sizes of the integer types in C: The value ranges of integer types can be found in the limits.h header file. This header file contains the macros that define minimum and maximum values of each integer type e.g., INT_MIN, INT_MAX for minimum and maximum size of the integer.

2,147,483,647 should be the correct value.

**What is the data type that occupies the least storage space that can ,** of memory for storage and can contain both positive and negative values. It is not the size of the integer that is the problem, it's the size of a stack-allocated array - the variable arr is about 2MB large, which probably blows the stack. Either use a global/static variable, or use dynyamic memory allocation. And to answer your actual question, the largest integer value is about 2147 million.

Your compiler is absolutely correct by informing you about the overflow.

To make it a bit simpler imagine a 4 bit integer. A four bit integer can hold up to 2^4=16 different values. Each value can only be a combination of these four bits:

8 4 2 1 0 0 0 0 = 0*8 + 0*4 + 0*2 + 0*1 = 0 0 0 0 1 = 0*8 + 0*4 + 0*2 + 1*1 = 1 0 0 1 0 = 0*8 + 0*4 + 1*2 + 0*1 = 2 ... 1 1 0 0 = 1*8 + 1*4 + 0*2 + 0*1 = 12 1 1 0 1 = 1*8 + 1*4 + 0*2 + 1*1 = 13 1 1 1 0 = 1*8 + 1*4 + 1*2 + 0*1 = 14 1 1 1 1 = 1*8 + 1*4 + 1*2 + 1*1 = 15

So therefore a four bit integer can have 16 different values but the largest value is 15 (= 16 - 1 = 2^4 - 1).

Back to a four byte integer which is made up of 32 bits. Same game as before: The 32 bits can be combined in 2^32 = 4294967296 different ways. The largest number that can be written by 32 bits is 2^32 - 1 = 4294967295.

To write a binary number of value 4294967296 you need an additional 33rd bit. Since your integer is only 32 bits long the compiler informs you about that.

**about the largest value an int can hold in c programming?,** You do not need to remember such constants by heart or calculate them yourself. The largest int value is defined in the file limits.h . The file also INT_MAX and INT_MIN in C/C++ and Applications Most of the times, in competitive programming, there is a need to assign the variable, the maximum or minimum value that data type can hold, but remembering such a large and precise number comes out to be a difficult job.

**C and C++ Integer Limits,** Maximum value for a variable of type char. 127; 255 if /J option used. MB_LEN_MAX, Maximum number of bytes in a multicharacter constant. 5. Defines the value for type char and its value will be equal to SCHAR_MAX if char represents negative values, otherwise UCHAR_MAX. Defines the maximum number of bytes in a multi-byte character. Defines the minimum value for a short int. Defines the maximum value for a short int.

**What is the max value integer can hold in C++?,** By using our site, you acknowledge that you have read and understood our Cookie INT_MIN specifies that an integer variable cannot store any value below this limit. C++ program to print values of INT_MAX int main(). {. // The sum of these numbers will equal INT_MAX. // If any of them is Assigning highest value. int In C, signed and unsigned are type modifiers. You can alter the data storage of a data type by using them. For example, unsigned int x; Here, the variable x can hold only zero and positive values because we have used the unsigned modifier. Considering the size of int is 4 bytes, variable y can hold values from -2 31 to 2 31 -1, whereas variable

**C Library - <limits.h>,** How do I write a C++ program for adding 2 numbers? 1,459 Views A 32 bit system will most likely have 32 bit integers, so the largest value is 4,294,967,296. In Python, value of an integer is not restricted by the number of bits and can expand to the limit of the available memory (Sources : this and this). Thus we never need any special arrangement for storing large numbers (Imagine doing above arithmetic in C/C++).

##### Comments

- Hint: is your
`int`

signed or unsigned? - it is signed one
- @praveenpadala - what makes you think that the largest value an
`int`

can hold is 4294967296? And if, as you've said, you're using a signed`int`

, then what is the smallest value a signed`int`

can hold? Please edit your question and walk us through your logic for that. Thanks. - Be careful when developing portability code, there are still MCUs supporting 16-bit signed integer. If so, using the typedefs of <types.h> with
`int32_t`

or`uint32_t`

as well as`int16_t`

or`uint16_t`

will help. - Note that
`4294967296`

is a 33-bit number. Unclear why one would think this is a maximal`int`

. - I forgot to mention it is signed one but it is not showing any warning up to the value 4294967295, even though it is signed int, why it is showing warning for only the value 4294967296 and above that value
- @praveenpadala: If you are actually getting no warning for
`int a = 4294967295;`

but are getting a warning for`int a = 4294967296;`

, that may be an interesting question. You should enter a question about that showing complete source code for each case and stating the full compiler name, version, and command-line switches. - @Eric: gcc only warns about
`int a = 4294967295`

if you specify`-Wconversion`

or`-pedantic`

. It warns about`int a = 4294967296`

if you specify`-Woverflow`

, which is included in`-Wall`

. I'm not trying to justify or anything :-). Clang warns on both, at least in the versions I have installed.