[] equivalence?

[Paul Williams]

I've got some char arrays declared, some as

char anArray[10] ;

and some as

char aChar ;

Im trying to ensure that all are null terminated, so am using:

anArray[0] = '\0' ;
aChar[0] = '\0' ;

But the second one does not compile - I get the error:

'The array operator must have one operand that is a pointer to
acomplete type and one of integral type'

I thought that aChar[n] was equivalent to (&aChar + n)?

Obviously aChar = '\0' ; works, but is that not equivalent to
aChar[0] = '\0' ;

ta

 

[Sharad Kala]

I've got some char arrays declared, some as

char anArray[10] ;

and some as

char aChar ;

Im trying to ensure that all are null terminated, so am using:

anArray[0] = '\0' ;
aChar[0] = '\0' ;

But the second one does not compile - I get the error:

'The array operator must have one operand that is a pointer to
acomplete type and one of integral type'

I thought that aChar[n] was equivalent to (&aChar + n)?

No, it's *(aChar + n)

Obviously aChar = '\0' ; works, but is that not equivalent to
aChar[0] = '\0' ;

Obviously, aChar[0] == *(aChar + 0), aChar is no pointer. Hence the error.
btw, there are array to pointer conversions. The result is a pointer to the
first element of the array.
Hence, anArray[0] == *( anArray + 0), here anArray is &anArray[0].

-Sharad

 

[Andre Heinen]

I've got some char arrays declared, some as
char anArray[10] ;
and some as
char aChar ;

This is *not* an array of char. It is just a char. A single
one.

<snip>
Obviously aChar = '\0' ; works,

Yes, because aChar is a char.

but is that not equivalent to
aChar[0] = '\0' ;

As aChar is not an array, you can't write aChar[0].

 

[Bill Seurer]

I've got some char arrays declared, some as

char anArray[10] ;

and some as

char aChar ;

That is not an array.

Im trying to ensure that all are null terminated, so am using:

anArray[0] = '\0' ;
aChar[0] = '\0' ;

But the second one does not compile - I get the error:

'The array operator must have one operand that is a pointer to
acomplete type and one of integral type'

I thought that aChar[n] was equivalent to (&aChar + n)?

No, arr[n] is equivalent to *(arr+n).

In any case aChar is not an array or pointer so you cannot using the
subscript operator on it. Why are you trying to do it this way anyway?

 

[Richard Herring]

I've got some char arrays declared, some as

char anArray[10] ;

and some as

char aChar ;

That's not an array.

Im trying to ensure that all are null terminated, so am using:

anArray[0] = '\0' ;
aChar[0] = '\0' ;

But the second one does not compile - I get the error:

'The array operator must have one operand that is a pointer to
acomplete type and one of integral type'

I thought that aChar[n] was equivalent to (&aChar + n)?

No, because aChar isn't of array type.

An lvalue or rvalue of array type can be converted to a pointer, so
anArray[n] is equivalent to * (&anArray[0] + n). Note the difference
from what you wrote above.

That conversion is only possible for arrays.

Obviously aChar = '\0' ; works, but is that not equivalent to
aChar[0] = '\0' ;