python - xor each byte with 0x71 -

i needed read byte file, xor 0x71 , write file. however, when use following, reads byte string, xoring creates problems.

f = open('a.out', 'r') f.read(1) 

so ended doing same in c.

#include <stdio.h> int main() {   char buffer[1] = {0};   file *fp = fopen("blah", "rb");   file *gp = fopen("a.out", "wb");   if(fp==null) printf("error opening file\n");   int rc;   while((rc = fgetc(fp))!=eof) {     printf("%x", rc ^ 0x71);     fputc(rc ^ 0x71, gp);   }   return 0; } 

could tell me how convert string on using f.read() on hex value xor 0x71 , subsequently write on file?

if want treat array of bytes, want bytearray behaves mutable array of bytes:

b = bytearray(open('a.out', 'rb').read()) in range(len(b)):     b[i] ^= 0x71 open('b.out', 'wb').write(b) 

indexing byte array returns integer between 0x00 , 0xff, , modifying in place avoid need create list , join again. note file opened binary ('rb') - in example use 'r' isn't idea.