1// RUN: %clang_cc1 -fsyntax-only -verify %s 2 3#define nil ((void*) 0) 4 5@interface A 6@property int x; 7@end 8 9@interface B : A 10@end 11 12// Basic checks... 13id f0(int cond, id a, void *b) { 14 return cond ? a : b; 15} 16A *f0_a(int cond, A *a, void *b) { 17 return cond ? a : b; 18} 19 20id f1(int cond, id a) { 21 return cond ? a : nil; 22} 23A *f1_a(int cond, A *a) { 24 return cond ? a : nil; 25} 26 27void *f1_const_a(int x, void *p, const A * q) { 28 void *r = x ? p : q; // expected-warning{{initializing 'void *' with an expression of type 'const void *' discards qualifiers}} 29 return r; 30} 31 32// Check interaction with qualified id 33 34@protocol P0 @end 35 36id f2(int cond, id<P0> a, void *b) { 37 return cond ? a : b; 38} 39 40id f3(int cond, id<P0> a) { 41 return cond ? a : nil; 42} 43 44// Check that result actually has correct type. 45 46// Using properties is one way to find the compiler internal type of a 47// conditional expression. Simple assignment doesn't work because if 48// the type is id then it can be implicitly promoted. 49@protocol P1 50@property int x; 51@end 52 53int f5(int cond, id<P1> a, id<P1> b) { 54 return (cond ? a : b).x; 55} 56int f5_a(int cond, A *a, A *b) { 57 return (cond ? a : b).x; 58} 59int f5_b(int cond, A *a, B *b) { 60 return (cond ? a : b).x; 61} 62 63int f6(int cond, id<P1> a, void *b) { 64 // This should result in something with id type, currently. 65 return (cond ? a : b).x; // expected-error {{member reference base type 'void *' is not a structure or union}} 66} 67 68int f7(int cond, id<P1> a) { 69 return (cond ? a : nil).x; 70} 71 72int f8(int cond, id<P1> a, A *b) { 73 return a == b; // expected-warning {{comparison of distinct pointer types ('id<P1>' and 'A *')}} 74} 75 76int f9(int cond, id<P1> a, A *b) { 77 return (cond ? a : b).x; // expected-warning {{incompatible operand types ('id<P1>' and 'A *')}} \ 78 expected-error {{property 'x' not found on object of type 'id'}} 79} 80