There is no namespace lookup order in C. Only one namespace is ever considered for any particular identifier; it is determined by what kind of identifier is being looked up. Structure tags are one kind, with their own namespace; variable names are in the broader category of "ordinary identifiers", which have a separate namespace. There are other name spaces, too, but the compiler can always tell from context which one is relevant to any given identifier.
Thus, in your program, both uses of myst.min refer to the variable declared as st myst; and there isn't any second variable in the "tag" namespace (as you seem to have thought there would be).
You can see this for yourself by commenting out everything within main above myst.min = 7:
#include <stdio.h>
int main (void){
#if 0
typedef struct{ //This belongs to ordinary name space
int min;
} st;
st myst;
myst.min=6;
#endif
struct myst{ // This belongs to tag name space
int min;
};
myst.min=7;
printf("%d\n%d\n",myst.min,myst.min);
return 0;
}
Attempting to compile this will produce a hard error:
test.c: In function ‘main’:
test.c:14:3: error: ‘myst’ undeclared (first use in this function)
What the struct myst declaration does is declare another type, which you can then use to declare variables. But it's not used for anything unless you actually do that. For instance
#include <stdio.h>
typedef struct { int min; } st;
struct myst { int min };
int main(void)
{
// uses the typedef name 'st', in the ordinary namespace,
// to declare the variable 'myst', also in the ordinary namespace
st myst = { 6 };
// uses the struct name 'myst', in the tag namespace,
// to declare the variable 'myst2', in the ordinary namespace
struct myst myst2 = { 7 };
printf("%d %d\n", myst.min, myst2.min);
return 0;
}
will print 6 7. This program also illustrates how myst, the variable, and myst, the struct tag, are indeed in two different namespaces and can be referred to independently.
(Thanks to John Bollinger for the new first paragraph. —ed)
