I'm solid confused with range of data type in C. I have 64 bit computer so int data type can store 64 bit value

64 bits: maximum representable value = 18,446,744,073,709,551,615

https://en.wikipedia.org/wiki/Integer_overflow

#include <stdio.h>

int main (void)
{
    int value  =  18446744073709551615;
  
    printf ("%d", value);
  
    return 0;
}

When I run program on 64 bit computer. It gives two warning

warning: integer constant is so large that it is unsigned
Warning: overflow in implicit constant conversion [-Woverflow]

Result : -1

Program should be print this value = 18446744073709551615 but I don't understand why it print -1.

How maximum value int data type can be hold in memory location ?

 

Change data type to unsigned int.

 

Change data type to unsigned int.

I have been changed data type but program gives -1 output with two warnings

 

"%u"

 

gcc -Wconversion -Wsign-conversion -Werror -Wall -Wextra -pedantic -o signwarn signwarn.c
signwarn.c: In function ‘main’:
signwarn.c:8:19: error: integer constant is so large that it is unsigned [-Werror]
int value = 18446744073709551615;
^~~~~~~~~~~~~~~~~~~~
signwarn.c:8:19: error: overflow in conversion from ‘__int128’ to ‘int’ changes value from ‘18446744073709551615’ to ‘-1’ [-Werror=overflow]

 

Or printf("%ld");

 

Result : -1

Program should be print this value = 18446744073709551615 but I don't understand why it print -1.

How maximum value int data type can be hold in memory location ?

You assumed that int has 8-byte storage, but clearly it does not. Try this:

    printf( "sizeof(short): %lu\n"
            "sizeof(int): %lu\n"
            "sizeof(long): %lu\n"
            "sizeof(long long): %lu\n",
            sizeof(short), sizeof(int), sizeof(long),
            sizeof(long long) );

The results will be in units of bytes. The C standard only guarantees the following:

\( \small{\text{sizeof(short) \le sizeof(int) \le sizeof(long) \le sizeof(long long)}}\)

Note that, regardless of size, in two's complement encoding the most-significant bit of signed types is a sign bit. This means that half of the possible encodings are negative values. To get the full positive range use an unsigned type.

 

Try with a lower until you reach the right one. Shoul be Value - 1.

 

"%u"

If I use %u then It print 4294967295 is not original value

Or printf("%ld");

If I use %ld it print -1

gcc -Wconversion -Wsign-
signwarn.c:8:19: error: overflow in conversion from ‘__int128’ to ‘int’ changes value from ‘18446744073709551615’ to ‘-1’ [-Werror=overflow]

Is this reason not being printed to the correct value.

 

Oct 1777777777777777777777
Bin 1111111111111111111111111111111111111111111111111111111111111111
Dec -1
Hex FFFFFFFFFFFFFFFF

Is the proper representation.

 

Try with a lower until you reach the right one. Shoul be Value - 1.

It only print up to 10 digit value for example : 1844674407

if value is 11 digits for example :18446744070

int value = 18446744070;

It print 1266874886

 

I'm solid confused with range of data type in C. I have 64 bit computer so int data type can store 64 bit value

Check the definitions in your header files.

This is from a 32 bit ARM:

/* Unsigned.  */
typedef unsigned char           uint8_t;
typedef unsigned short int      uint16_t;
#ifndef __uint32_t_defined
typedef unsigned int            uint32_t;
# define __uint32_t_defined
#endif
#if __WORDSIZE == 64
typedef unsigned long int       uint64_t;
#else
__extension__
typedef unsigned long long int  uint64_t;
#endif

On a 64 bit ARM architecture, I'd have to use unsigned long int to get 8 bytes.

 

If you google printf() you will see whats needed..

d or I signed int ( 16 bits )
u unsigned int ( 16 bits )
ld ( long signed . 32 bits)
lu ( long unsigned . 32 bits )

lld ( long long signed . 64 bits )
llu ( long long unsigned . 64 bits )

You choose...​

 

I'm solid confused with range of data type in C. I have 64 bit computer so int data type can store 64 bit value

64 bits: maximum representable value = 18,446,744,073,709,551,615

https://en.wikipedia.org/wiki/Integer_overflow

#include <stdio.h>

int main (void)
{
    int value  =  18446744073709551615;
 
    printf ("%d", value);
 
    return 0;
}

When I run program on 64 bit computer. It gives two warning

warning: integer constant is so large that it is unsigned
Warning: overflow in implicit constant conversion [-Woverflow]

Result : -1

Program should be print this value = 18446744073709551615 but I don't understand why it print -1.

How maximum value int data type can be hold in memory location ?

Your computer is 64 bits but what about your compiler?

Picbuster

 

Is this reason not being printed to the correct value.

You can use a GCC compiler __int128 variable to hold the constant (that's limited to a 64-bit value), side-step the 64 bit sign conversion in a 64-bit int and then print value.

# cat signwarn.c

#include <stdint.h>
#include <stdio.h>

int main(void) {
__int128 value = 18446744073709551615;

printf ("\r\n%llu\r\n", (long long int) value);
return 0;
}

# gcc -Wall -Wextra -o signwarn signwarn.c
signwarn.c: In function ‘main’:
signwarn.c:5:24: warning: integer constant is so large that it is unsigned
__int128 value = 18446744073709551615;
^~~~~~~~~~~~~~~~~~~~
# ./signwarn

18446744073709551615

 

On x64, using gcc:

#include <stdio.h>

int main (void)
{
  unsigned long value  =  18446744073709551615UL; // note the 'UL' suffix

  printf( "value: %lu\n", value );

  return 0;
}

value: 18446744073709551615

 

Your computer is 64 bits but what about your compiler?

Picbuster

I had downloaded compiler from this website https://sourceforge.net/projects/mingw-w64/ It supports 32-bit and 64-bit Windows

@nsaspook , @bogosort I have run both of your program on my computer It print value 4294967295

I wonder why does it print 4294967295 on my computer

 

I had downloaded compiler from this website https://sourceforge.net/projects/mingw-w64/ It supports 32-bit and 64-bit Windows

@nsaspook , @bogosort I have run both of your program on my computer It print value 4294967295

I wonder why does it print 4294967295 on my computer

My version __int128 won't work in gcc -m32 32-bit mode but this is what I get with the @bogosort code.

# gcc -Wall -Wextra -m32 -o signwarn signwarn.c
signwarn.c: In function ‘main’:
signwarn.c:9:27: warning: conversion from ‘long long unsigned int’ to ‘long unsigned int’ changes value from ‘18446744073709551615’ to ‘4294967295’ [-Woverflow]
unsigned long value = 18446744073709551615UL; // note the 'UL' suffix
^~~~~~~~~~~~~~~~~~~~~~
# ./signwarn
value: 4294967295
# gcc -Wall -Wextra -m64 -o signwarn signwarn.c
# ./signwarn
value: 18446744073709551615

Seems like your compiler was in 32-bit mode even if it runs on 64-bit windows.

 

Seems like your compiler was in 32-bit mode even if it runs on 64-bit windows.

I think you're right

#include <stdio.h>

int main (void)
{
     unsigned long value  =  4294967295 ;

     printf( "value: %lu\n", value );

    return 0;
}

This program print value: 4294967295

my original question was How maximum value int data type can be hold in memory location ?

consider three case

int value : How maximum value it can be hold in memory location ?

unsigned int value : How maximum value it can be hold in memory location ?

signed int value : How maximum value it can be hold in memory location ?

 

I think you're right

#include <stdio.h>

int main (void)
{
     unsigned long value  =  4294967295 ;

     printf( "value: %lu\n", value );

    return 0;
}

This program print value: 4294967295

my original question was How maximum value int data type can be hold in memory location ?

consider three case

int value : How maximum value it can be hold in memory location ?

unsigned int value : How maximum value it can be hold in memory location ?

signed int value : How maximum value it can be hold in memory location ?

All the information you need is in: https://en.wikibooks.org/wiki/C_Programming/stdint.h