Yes: use the z length modifier: size_t size = sizeof(char); printf(“the size is %zu\n”, size); // decimal size_t (“u” for unsigned) printf(“the size is %zx\n”, size); // hex size_t The other length modifiers that are available are hh (for char), h (for short), l (for long), ll (for long long), j (for intmax_t), t (for … Read more
You may be able to exploit a format string vulnerability in many ways, directly or indirectly. Let’s use the following as an example (assuming no relevant OS protections, which is very rare anyways): int main(int argc, char **argv) { char text[1024]; static int some_value = -72; strcpy(text, argv[1]); /* ignore the buffer overflow here */ … Read more
Most of these answers explain what %n does (which is to print nothing and to write the number of characters printed thus far to an int variable), but so far no one has really given an example of what use it has. Here is one: int n; printf(“%s: %nFoo\n”, “hello”, &n); printf(“%*sBar\n”, n, “”); will … Read more
You can escape it by posting a double ‘%’ like this: %% Using your example: printf(“hello%%”); Escaping the ‘%’ sign is only for printf. If you do: char a[5]; strcpy(a, “%%”); printf(“This is a’s value: %s\n”, a); It will print: This is a’s value: %%