sizeof(int) is 4 bytes but only two are written
c, integer, sizeof
Solution
You have written text to the file. That is what `fprintf` does. It converts your parameters into text according to your format string, and then puts that text to the file. You have actually written three bytes to the file: two ASCII digits ('1' and '4') and a line feed ('\n'). Note that 0x31 is the ASCII code for `'1'` and 0x34 is the ASCII code for `'4'`, and 0x0a is the ASCII code for a line feed.
You need to write binary instead. Use `fwrite` for that.
int i = 14;
fwrite(&i, sizeof(i), 1, fp);
Problem
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 gives the following output to `stdout`: ``` sizeof(int)=4 ``` test.bin has the following contents when viewed in any text editor: ``` 14 ``` When viewed with `vi` using the hex dump option (`xxd`): ``` 0000000: 3134 0a 14. ``` When viewed with `hexdump -c`: ``` 0000000 1 4 \n 0000003 ``` Obviously an integer on my machine is four bytes, but both `hexdump` and `vi` are telling me that only two bytes were required to represent 14, and another byte was used to represent the newline character. This is confirmed by the fact that test.bin is only three bytes in size. But, an integer is four bytes, so why are only two bytes representing it? What obvious fact have I completely forgotten and cant seem to remember?