Why const variable can not define array bounds.

[MartinR]

Hello,

I found out that below code is illegal:

const int a = 10;
int b[a];

So, why second line is incorrect? After all content of a cannot be change since it is declared as 'const'. Second even if it wasn't const, they memory for array shall be reserved once, so who cares what happened to variable a (in this example) after memory for array was allocated?

[laserlight]

So, why second line is incorrect?

Basically just the rules of C concerning what exactly is a constant expression in this context, e.g., an integer constant like 10 is a constant expression, but an int variable like a, even if declared const, is not. Consequently, one way out is to use a macro to define the constant, since macro replacement would cause the identifier to be replaced by the integer constant during preprocessing.

After all content of a cannot be change since it is declared as 'const'.

True, but it can be more complicated, e.g.,

const int a = c;
int b[a];

If c is not const, then at run time it is possible for a to have varying values on each run (or each call of the enclosing function, etc). So disallowing this or saying that b is therefore a variable length array simplifies the situation. There are ways to deal with these more complex situations, of course, but standard C did not go down that path.

Second even if it wasn't const, they memory for array shall be reserved once, so who cares what happened to variable a (in this example) after memory for array was allocated?

Because the amount of memory to allocate for the array may be determined at compile time. What you have in mind is more of a variable length array, where the amount of memory to allocate may be determined at run time, but cannot change after being allocated. Therefore, it is necessary to determine the value at compile time, hence the rule about it being a constant expression.

[MartinR]

Got it! Thanks