Initialize Global Variable With Function Call Assignment

Question

I have a simple program as below:

static const char* DeviceID = (char*)"my id";
int length = strlen(DeviceID);
int main(){

}

and the compiler throw the following error:

initializer element is not constant

I don't know why the compiler can't understand my statement:
strlen's prototype is like the following code:

size_t strlen ( const char * str );

Because it is outside a function? C doesn't have static constructors; functions can only be called from within functions. Call strlen where you need it. Or are you using C++? (You specify both tags.) — M Oehm
– M Oehm, Commented Jun 3, 2014 at 9:26
Why did you tag this C++ if you're not using a C++ compiler? — user743382
– user743382, Commented Jun 3, 2014 at 9:27
Also, in this particular case you don't even need to call an actual function, if you do static const char[] DeviceID = "my id"; you can use sizeof on it. — filmor
– filmor, Commented Jun 3, 2014 at 9:28

pmg · Accepted Answer · 2014-06-03 09:29:43Z

1

Try sizeof which yields a compile time constant

#define MY_ID "my id"
static const char *DeviceID = MY_ID;  // no cast needed
int length = sizeof MY_ID - 1;        // sizeof also includes the '\0'

int main(void) {
    /* ... */
}

edited Jun 3, 2014 at 9:29

answered Jun 3, 2014 at 9:28

pmg

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Comments

kniz · Accepted Answer · 2016-06-03 07:06:46Z

1

As far as I know, there is different with C and C++ when you're trying to initailize global variables.

in C, Rvalue of the global initializing statement should be evaluated at compile time.
for instance,

static const char* DeviceID = (char*)"my id"; <-- compiler can evaluate address of "my_id". so, this is valid.

int length = strlen(DeviceID); <-- but this can't be at compiletime. means that, we should run the process in order to know its value.

But, C++ doesn't care about like above. it can initialize global ones at runtime. so if you use C++ compiler instead of, errors'll be disappeared.

answered Jun 3, 2016 at 7:06

kniz

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