How to specify 64 bit integers in c

2019-01-16 18:39发布

问题:

I'm trying to use 64 bit integers in C, but am getting mixed signals as to whether it should be possible.

When I execute the printf:

printf("Size of long int:%d\nSize of long long int:%d\n\n",(int)sizeof(long int), (int)sizeof(long long int));

The response I get is:

Size of long int:4 Size of long long int:8

This makes me feel that a long long int has 8 bytes = 64 bits.

However, when I try to declare the following variables:

long long int a2 = 0x00004444;
long long int b2 = 0x000044440;
long long int c2 = 0x0000444400;
long long int d2 = 0x00004444000;
long long int e2 = 0x000044440000;
long long int f2 = 0x0000444400004;
long long int g2 = 0x00004444000044;
long long int h2 = 0x000044440000444;
long long int i2 = 0x0000444400004444;

The last 4 variables (f2,g2,h2,i2) give me the error message:

warning: integer constant is too large for ‘long’ type

I get the same result when I replace 'long long int' with 'int64_t'. I assume 'int64_t' was recognized, since it didn't generate any error messages of its own.

So, it appears my 8 byte long long int is really a 6 byte long long int, and I don't understand what I'm missing here. If it's any help, here is the information on my gcc compiler:

me@ubuntu:~$ gcc -v  
Using built-in specs.  
Target: i686-linux-gnu  
Configured with: ../src/configure -v   
--with-pkgversion='Ubuntu/Linaro 4.4.4-14ubuntu5'  
--with-bugurl=file:///usr/share/doc/gcc-4.4/README.Bugs   
--enable-languages=c,c++,fortran,objc,obj-c++  
--prefix=/usr   
--program-suffix=-4.4   
--enable-shared   
--enable-multiarch   
--enable-linker-build-id   
--with-system-zlib   
--libexecdir=/usr/lib   
--without-included-gettext   
--enable-threads=posix   
--with-gxx-include-dir=/usr/include/c++/4.4   
--libdir=/usr/lib   
--enable-nls   
--with-sysroot=/ -  
-enable-clocale=gnu   
--enable-libstdcxx-debug   
--enable-objc-gc   
--enable-targets=all 
--disable-werror   
--with-arch-32=i686   
--with-tune=generic   
--enable-checking=release   
--build=i686-linux-gnu   
--host=i686-linux-gnu   
--target=i686-linux-gnu  
Thread model: posix  
gcc version 4.4.5 (Ubuntu/Linaro 4.4.4-14ubuntu5)   

If anyone knows how (or if) 64 bit integers are accessible to me, I'd really appreciate any help. Thanks....

回答1:

Use stdint.h for specific sizes of integer data types, and also use appropriate suffixes for integer literal constants, e.g.:

#include <stdint.h>

int64_t i2 = 0x0000444400004444LL;


回答2:

Try an LL suffix on the number, the compiler may be casting it to an intermediate type as part of the parse. See http://gcc.gnu.org/onlinedocs/gcc/Long-Long.html

long long int i2 = 0x0000444400004444LL;

Additionally, the the compiler is discarding the leading zeros, so 0x000044440000 is becoming 0x44440000, which is a perfectly acceptable 32-bit integer (which is why you aren't seeing any warnings prior to f2).



回答3:

Use int64_t, that portable C99 code.

int64_t var = 0x0000444400004444LL;

For printing:

#define __STDC_FORMAT_MACROS
#include <inttypes.h>

printf("blabla %" PRIi64 " blabla\n", var);


回答4:

How to specify 64 bit integers in c

Going against the usual good idea to appending LL.

Appending LL to a integer constant will insure the type is at least as wide as long long. If the integer constant is octal or hex, the constant will become unsigned long long if needed.

If ones does not care to specify too wide a type, then LL is OK. else, read on.

long long may be wider than 64-bit.

Today, it is rare that long long is not 64-bit, yet C specifies long long to be at least 64-bit. So by using LL, in the future, code may be specifying, say, a 128-bit number.

C has Macros for integer constants which in the below case will be type int_least64_t

#include <stdint.h>
#include <inttypes.h>

int main(void) {
  int64_t big = INT64_C(9223372036854775807);
  printf("%" PRId64 "\n", big);
  uint64_t jenny = INT64_C(0x08675309) << 32;  // shift was done on at least 64-bit type 
  printf("0x%" PRIX64 "\n", jenny);
}

output

9223372036854775807
0x867530900000000


回答5:

Append ll suffix to hex digits for 64-bit (long long int), or ull suffix for unsigned 64-bit (unsigned long long)



标签: c 64bit sizeof