CODEWITHSUNDEEP
c
prap19
prap19

Reputation: 1868

Why this program is truncating output?

Am I doing anything wrong in this following program?

Code

    #include <stdio.h>
int main()
{
long x=1290323123123;
int len = snprintf(NULL,0, "%ld", x);

printf("%ld  %ld",x,len);

 return 0;
}

Output: 1832934323 10

Upvotes: 1

Views: 114

Answers (2)

kennytm
kennytm

Reputation: 523314

1290323123123 requires 41 bits to store, but that long probably is just 32-bit long, so the extra 9 bits are gone.

1290323123123 = 0x12c6d405bb3
                  ^^^
                  excessive data that is chopped off

              =    0x6d405bb3

              = 1832934323

Use

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

int main () {
    int64_t x = 1290323123123LL;
//  ^^^^^^^                  ^^
    int len = snprintf(NULL, 0, "%"PRId64, x);
//                                ^^^^^^^
    printf("%"PRId64" %d\n", x, len);
//           ^^^^^^^^
    return 0;
}

to make sure the type is at least 64-bit long, so it can store that value completely (result: http://www.ideone.com/BnTjJ).

Upvotes: 6

Sam Skuce
Sam Skuce

Reputation: 1694

Your 'long' type can only hold 4 bytes. The value you've assigned 'x' is greater than 4 bytes. 

Hex(1290323123123) = 12C 6D40 5BB3
Hex(1832934323 ) = 6D40 5BB3

So the number it's outputting is the same as the lower 4 bytes of the number you're trying to print.

Some compilers may have larger 'long' types - prior to C99 and the introduction of types like int64_t I don't think there is one standard for (name, size) pairs.

Upvotes: 2

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