How to convert decimal to binary in 64 bits? - c

Question

so I have this code
int main()
{
int n, c, k;
printf("Enter an integer\n");
scanf("%d", &n);
printf("%d in binary is:\n", n);
for (c = 31; c >= 0; c--)
{
k = n >> c;
if (k & 1)
printf("1");
else
printf("0");
}
printf("\n");
return 0;
}
It converts decimal into binary but only in 32 bits. When I change it into 64 bits it doesn't work (it seems like it just doubles the result from 32 bits). At the same time it works fine with 8 or 4 bits etc.
What am I doing wrong?

Answer 1

It converts decimal into binary but only in 32 bits. When I change it into 64 bits it doesn't work (it seems like it just doubles the result from 32 bits).
The problem is here.
int n, c, k;
printf("Enter an integer\n");
scanf("%d", &n);
n is an int which can be as small as 16 bits. It could be 64 bits, but it's probably 32 bits. When you try to enter a 64 bit number you'll get garbage.
#include <stdio.h>
int main() {
int n;
printf("sizeof(int) == %zu\n", sizeof(int));
printf("Enter an integer\n");
scanf("%d", &n);
printf("n = %d\n", n);
}
$ ./test
sizeof(int) == 4
Enter an integer
12345678901
n = -539222987
Instead, you can use a long long int which has a minimum size of 64 bits or int64_t from stdint.h which is exactly 64 bits. I have a preference for using the explicit width types in code that requires a specific width to make it more obvious to the reader.
long long int uses %lld for scanf and printf while int64_t uses macros from inttypes.h. The code below takes advantage of C automatically concatenating constant strings; "foo" "bar" and "foobar" are equivalent.
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main() {
int64_t n;
printf("Enter an integer\n");
scanf("%"SCNd64, &n);
printf("n = %"PRId64"\n", n);
}
$ ./test
Enter an integer
12345678901
n = 12345678901

Answer 2

The problem is your n & k variables' data type. They are integer. Check your platform's data type size using sizeof(int).
When you change to 64-bits, these variables can't hold 64-bit values.
HTH!

Answer 3

the following code is one way to handle the problem:
#include <stdio.h>
#include <stdlib.h>
int main( void )
{
long long unsigned int n;
long long int c;
long long unsigned int k;
printf("Enter an integer\n");
if( 1 != scanf("%llu", &n) )
{
perror( "scanf for 64 bit number failed" );
exit( EXIT_FAILURE );
}
// implied else, scanf successful
printf("%llu in binary is:\n", n);
for (c = 63; c >= 0; c--)
{
k = n >> c;
if (k & 1)
putc('1', stdout);
else
putc('0', stdout);
}
printf("\n");
return 0;
} // end function: main
and here is the output from a typical run of the program:
Enter an integer
6789097860397846
6789097860397846 in binary is:
0000000000011000000111101010011000000110010100000111101100010110