Why can't static variables be initialized with values of other variables?

Question

What is the reason for this?

const int a = 0;
static int b = a * 5;  // compile error
int main()
{
    const int x = 1;
    static int y = x * 10;  // compile error
}

http://stackoverflow.com/a/3025106/257418 – Lynn Apr 12 '15 at 00:56 — Lynn, Apr 12 '15 at 00:56
`a` also isn't a "constant", it's a "variable". – BLUEPIXY Apr 12 '15 at 00:57 — BLUEPIXY, Apr 12 '15 at 00:57

AlexD · Accepted Answer · 2015-04-12T01:04:21.190

2

According to C standard:

All the expressions in an initializer for an object that has static or thread storage duration shall be constant expressions or string literals.

This is valid C++ code though.

edited Apr 12 '15 at 01:04

answered Apr 12 '15 at 00:57

AlexD

2

Specifically `a` is not a constant expression, it actually is a variable. (It would be a "constant" in C++; this limitation is why people use `#define` a lot in C.) – Lynn Apr 12 '15 at 00:58

1 Answers1