Why can I assign items to an array that has a length of 0?-CodePudding

So I have this code:

#include <stdio.h>
int main()
{
printf("enter character\n>>>");
char input[0];


scanf("%5s",input);
printf("%s",input);
}

that excepts 5 chars from the user. I am new to C and this one thing makes no sense to me. Why does gcc allow me to compile a program that assigns values to an array with a length of 0? Surely this is not possible? Please explain.

CodePudding user response：

Your compiler ought to reject the declaration as invalid

If the expression is a constant expression, it shall have a value greater than zero. (6.7.6.2).

However, as @Joshua points out below, some compilers support this feature as an extension:

Declaring zero-length arrays is allowed in GNU C as an extension (info gcc; 6.18)

gcc -pendatic -pedantic-errors will generate an error and only warning without -pedantic-errors.

scanf() and printf() will also be undefined behavior.

CodePudding user response：

C doesn't check for buffer overrun.

This bears repeating.

C doesn't check for buffer overrun.

This has been a source of bugs for a very long time; but also it's inherent in C and cannot be changed.

There are some simple cases where the compiler can detect buffer overrun (usually with optimizations enabled as well) but in the general case it cannot; nor will it generate any runtime checks. It will just do something unexpected. This is usually a security problem if you let it.

You must check yourself that you don't overrun buffers.

CodePudding user response：

This program causes undefined behaviour. Anything can happen, including the appearance of "correct behaviour." C is not the kind of language that can protect you from every potential memory management pitfall.