Why does this compile?

[g36130]

Hello!

I was wondering why the following C code compiles:

main()
{
int a[100];
int b= 4[a]; /* Looks like 4[a] is equivallent to a[4] ??? */
return b;
}

Any explanations/links that would help me understand this would be
welcome (I coundn't find that in the C standard).
Cheers
g

 

[Mark Bluemel]

Hello!

I was wondering why the following C code compiles:

main()
{
int a[100];
int b= 4[a]; /* Looks like 4[a] is equivallent to a[4] ??? */
return b;
}

Any explanations/links that would help me understand this would be
welcome (I coundn't find that in the C standard).

You could have found it in the C FAQ at http://www.c-faq.com/
It's question 6.11

 

[vippstar]

Hello!

I was wondering why the following C code compiles:

main()
{
int a[100];
int b= 4[a]; /* Looks like 4[a] is equivallent to a[4] ??? */
return b;

}

Any explanations/links that would help me understand this would be
welcome (I coundn't find that in the C standard).

You would better wonder why your code shouldn't compile.

main has to return int.
Evaluating uninitialized objects invokes undefined behavior.
As for your question, x[y] is equal to *(x + y), x+y == y+x.

 

[Malcolm McLean]

I was wondering why the following C code compiles:

main()
{
int a[100];
int b= 4[a]; /* Looks like 4[a] is equivallent to a[4] ??? */
return b;
}

Any explanations/links that would help me understand this would be
welcome (I coundn't find that in the C standard).

It's a silly quirk of the language.
Arrays can be treated as pointers in most contexts, and pointers as arrays.
a[4] = means *(a + 4). So it makes sense that 4[a] should mean *(4+a). But
to write that in code is rather pointless.

 

[Army1987]

Hello!

I was wondering why the following C code compiles:

main()
{
int a[100];
int b= 4[a]; /* Looks like 4[a] is equivallent to a[4] ??? */
return b;

}

You would better wonder why your code shouldn't compile.

main has to return int.

main() is equivalent to int main() in C90.

Evaluating uninitialized objects invokes undefined behavior.

This doesn't mean it shouldn't compile.

 

[Keith Thompson]

Arrays can be treated as pointers in most contexts, and pointers as arrays.

Um, sort of. Read section 6 of the comp.lang.c FAQ to understand what
this really means.