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Sizeof (int) - 4 bytes, but only two are written

Consider the following simple program:

#include <stdio.h>

int main () {
    FILE *fp;

    printf("sizeof(int)=%d\n", sizeof(int));

    fp = fopen("test.bin", "wb");
    fprintf(fp, "%d", 14);
    fclose(fp);
    return 0;
}

It outputs the following result to stdout:

sizeof(int)=4

test.bin has the following contents when viewed in any text editor:

14

When browsing with using the vihex dump ( xxd) option :

0000000: 3134 0a                                   14.

When viewing with hexdump -c:

0000000   1   4  \n
0000003

It is obvious that the integer on my machine is four bytes, but both hexdumpand vitell me to represent 14 only requires two bytes to represent a newline character used another byte. This is confirmed by the fact that test.bin has only three bytes. But the integer is four bytes, so why only two bytes represent it?

What obvious fact have I completely forgotten, and perhaps I don’t remember?

+4
3

. , fprintf. , . : ASCII ('1' '4') ('\ '). : 0x31 - ASCII '1', 0x34 - ASCII '4', 0x0a - ASCII .

. fwrite .

int i = 14;
fwrite(&i, sizeof(i), 1, fp);
+8
fprintf(fp, "%d", 14);

14 , 31 34 , ASCII '1' '4'.

+4

, . :

fprintf(fp, "%d", 14);

To:

int i = 14;
fwrite(&i, sizeof(int), 1, fp);

man page fwrite. , x86 AMD64, . , , , htonl <arpa/inet.h>. fread, , ntohl.

+1

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