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I am slightly confused about using "extern" in my c code, when there is a global variable involved. I tried the following, and got a compilation error:

Main.c:

extern unsigned short *videobuffer;
//I also tried this in a separate and it failed with the same compilation error//
extern (unsigned short *)videobuffer;

lib.c:

unsigned short *videobuffer = (unsigned short *)0x6000000;

The error I received: 

[COMPILE] Compiling main.c
main.c:16: error: expected identifier or '(' before 'unsigned'
make: *** [main.o] Error 1
Shanty
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  • How do you compile and link these two sources? You can *compile* but not *link* just your `main.c`. (And better don't try random stuff such as throwing in parentheses "until it works". You can easily look up the correct syntax elsewhere.) – Jongware Oct 24 '15 at 23:13

2 Answers2

0

Valid code is usually what people want here:

main.c

#include <stdio.h>
extern unsigned short *videobuffer;

int main() {
    printf("%p\n", videobuffer);
}

lib.c

unsigned short *videobuffer = (unsigned short *)0x6000000;
Bill Lynch
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0

the following is correct:

Main.c:

extern unsigned short *videobuffer;

lib.c:

unsigned short *videobuffer = (unsigned short *)0x6000000;

The extern keyword tells the compiler there is a variable, but it is not in this compilation unit.

Paul Ogilvie
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  • I tried this, but I am still getting the same error in compilation as stated above. – Shanty Oct 24 '15 at 22:22
  • Then you declare it at the wrong place. It must be before any statement and would best be declared before any function, after all includes. There can also be a syntax error before the declaration, which only triggers a compiler error later. Please show us more code. – Paul Ogilvie Oct 24 '15 at 22:41