Question

I have implemented below code :

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<cstdlib>
#include<sys/types.h>
main()
{
    int64_t i64value1 = 0;
    int64_t i64value2 = 0;
    long long llvalue = 0;
    const char *s = "10811535359";
    i64value1 = atoll(s);
    llvalue = atoll(s);
    i64value2 = llvalue;
    printf("s : [%s]\n",s);
    printf("i64value1 : [%d]\n",i64value1);
    printf("llvalue : [%lld]\n",llvalue);
    printf("i64value2 : [%d]\n",i64value2);
}

Output of the above progrom is :

s : [10811535359]
i64value1 : [-2073366529]
llvalue : [10811535359]
i64value2 : [-2073366529]

The compiler used is :

 gcc version 4.1.2 20080704 (Red Hat 4.1.2-48)

The OS is x86_64 GNU/Linux 2.6.18-194

Since long long is a signed 64-bit integer and is, for all intents and purposes, identical to int64_t type, logically int64_t and long long should be equivalent types. And some places mention to use int64_t instead of long long.
But when I look at stdint.h, it tells me why I see the above behavior:

# if __WORDSIZE == 64 
typedef long int  int64_t; 
# else 
__extension__ 
typedef long long int  int64_t; 
# endif 

In a 64-bit compile, int64_t is long int, not a long long int.

My Question is, Is there a workaround/solution to assign long long returned value to int64_t without losing the precision in 64 bit Machine?

Thanks in advance

Answer 1

The loss does not happen in the conversions but in the printing:

printf("i64value1 : [%d]\n",i64value1);

The int64_t argument is accessed as if it were an int. This is undefined behaviour, but usually the low 32 bits are sign extended.

Proper compiler warnings (such as gcc -Wformat) should complain about this.