c fgetpos giving wrong position

Question

I am new to C and I have this code:

f = fopen( argv[1], "rb" );
fseek( f, 64, SEEK_SET );
fpos_t pos;
fgetpos (f, &pos);
printf("%x", pos);

However, this returns 40, even though it's supposed to be returning 64. What am i doing wrong?

For one thing, is there actually 64 bytes in the file you're testing this with? — John Chadwick, May 05 '11 at 12:15

score 4 · Answer 1 · answered May 05 '11 at 12:17

4

You are outputting 64 in hex format, "%x". Since 64=0x40, the mystery is solved!

answered May 05 '11 at 12:17

David Heffernan

601,492
42
1,072
1,490

score 2 · Accepted Answer · answered May 05 '11 at 12:17

2

Because you're using %x. It's saying 40 as in 0x40, the hexadecimal number. You need %i or %d to get a decimal number.

answered May 05 '11 at 12:17

John Chadwick

3,193
19
19

score 1 · Answer 3 · answered May 05 '11 at 12:16

1

4 * 16 is 64

answered May 05 '11 at 12:16

pmg

106,608
13
126
198

score 1 · Answer 4 · answered May 05 '11 at 12:18

1

whatever you are printing is in hex format. 40 in decimal is 64. Do you mean the file size is 0x64 or 0x40

answered May 05 '11 at 12:18

Mayank

5,454
9
37
60

score 1 · Answer 5 · answered May 05 '11 at 12:34

fpos_t is not (necessarily) an arithmetic type and cannot be used with printf. An implementation could even store it as a structure containing an encrypted position if it liked. Use ftell (or ftello if available) to get the file offset in a meaningful numeric form. fgetpos is largely useless.

c fgetpos giving wrong position

5 Answers5