Want to XOR two strings fetched from argv. I checked this question How to xor two string in C? but it could not solve it for me.
#include <stdio.h>
#include <string.h>
int main(int argc, char const *argv[]) {
char output[]="";
int i;
for (i=0; i<strlen(argv[1]); i++){
char temp = argv[1][i]^argv[2][i];
output[i]= temp;
}
output[i] = '\0';
printf("XOR: %s\n",output);
return 0;
}
When I use lldb to debug my output ("(lldb) print output") it is /a/x16/t/x13 but it can not be printed by printf(). I know that it is not a string anymore. Can you help me how to make it able to be printf:ed. The text that is printed in the terminal is "XOR: "
There's some memory bugs in your code. Perhaps the following would work better:
#include <stdio.h>
#include <string.h>
#define min(i, j) ((i) < (j) ? (i) : (j))
int main(int argc, char const *argv[])
{
char *output;
int i;
/* Allocate a buffer large enough to hold the smallest of the two strings
* passed in, plus one byte for the trailing NUL required at the end of
* all strings.
*/
output = malloc(min(strlen(argv[1]), strlen(argv[2])) + 1);
/* Iterate through the strings, XORing bytes from each string together
* until the smallest string has been consumed. We can't go beyond the
* length of the smallest string without potentially causing a memory
* access error.
*/
for(i = 0; i < min(strlen(argv[1]), strlen(argv[2])) ; i++)
output[i] = argv[1][i] ^ argv[2][i];
/* Add a NUL character on the end of the generated string. This could
* equally well be written as
*
* output[min(strlen(argv[1]), strlen(argv[2]))] = 0;
*
* to demonstrate the intent of the code.
*/
output[i] = '\0';
/* Print the XORed string. Note that if characters in argv[1]
* and argv[2] with matching indexes are the same the resultant byte
* in the XORed result will be zero, which will terminate the string.
*/
printf("XOR: %s\n", output);
return 0;
}
As far as printf
goes, keep in mind that x ^ x
= 0 and that \0
is the string terminator in C.
Best of luck.