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extern is used outside.It should be used inside code.

When both file links during execution extern int *myarray declaration in file2.c will search for *myarray declared and allocated space ,and that is done in file1.c myarray[10].So what exactly you are saying is *myarray will not match to myarray[].

c-faq.com/decl/extarraysize.html

instead of extern int *myarray, extern int myarray[10] should be used and in file 2 we have to include file1 as header .which can be done by including #include at the front of source file i.e file 2 in this case.

after running i got
"error: conflicting types for ‘myarray’"

after running with gcc compiler,i got
"error: conflicting types for ‘myarray’"

arrays are special types which represents a continuous storage location . here you are making a pointer to an array , which would actually concider the first element i.e myarray [0] as its value
. when you deference *myarray it actually goes to the "myarray[0]" address which may be valid or invalid ..
i think linker should have complained i dont know ..
the confusion occurs because these statement works perfectly
int array[10];
int *ptr = array ;
ptr[0] = 10 ;

The code may give segmentation fault or linking error.

int array[10]; tells that array is an array of 10 integers.
extern int*myarray; tells that myarray is a pointer to an integer which in defined at some other, place probably in some other .c file. But it is not defined any where(even not initiallised).