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Why this code is unsecured?
#include <stdio.h>
int main( int argc, char *argv[] )
{
printf(argv[1]);
printf("\n");
return 0;
}
printfwill process its first parameter, looking for things like%dand%s.Based on those values, it will get more data from the stack and print it out.
So if someone called your program:
They could view a section of your computer’s callstack.
If they got even more creative with the format specifier, maybe they could dump something important, like a credit-card number or a password.