If there are insufficient arguments for the format, the behavior is undefined

I mistakenly wrote this code:

#include <stdio.h>

int main() {
    printf("%d");  // Missing argument!
    return 0;
}

Surprisingly, it compiled (warnings existed), but when I ran it, I got unexpected output or sometimes even a crash

• Why does printf("%d") compile successfully despite the missing argument?
• What happens when printf("%d") is executed without a corresponding integer?

1. printf is a variadic function

   printf takes a format string to interpret additional arguments. Since %d expects an integer, but none is provided, printf still tries to fetch an argument from the stack (or a register), leading to undefined behavior (UB).

2. Why does it compile?

   C does not check the number of arguments passed to printf() at compile time (unless warnings are enabled).

3. What happens at runtime?

   Since printf("%d") expects an integer but gets random data instead, the output is garbage or might cause a segmentation fault.

How to Fix it

Always provide the correct argument:

printf("%d", 42);  // Correct

Enable compiler warnings to catch such mistakes:

gcc -Wall -Wformat -o program program.c