Handle comma operator for implicit int->enum conversions #139536
Description
PR introduces a new diagnostic group to diagnose implicit casts from int to an enumeration type. In C, this is valid, but it is not compatible with C++.
However, in a case like the following
void free ();
typedef enum {REG_EBRACE, REG_BADBR} reg_errcode_t;
typedef struct {int *stack;} compile_stack_type;
reg_errcode_t byte_regex_compile () {
const char *p;
const char *pend;
compile_stack_type compile_stack;
return (free (compile_stack.stack), p == pend ? REG_EBRACE : REG_BADBR));
}
cmd: clang -Wimplicit-int-enum-cast -c pre.i
The warning is emitted.
pre.i:8:39: warning: implicit conversion from 'int' to enumeration type 'reg_errcode_t' is invalid in C++ [-Wimplicit-int-enum-cast]
8 | return (free (compile_stack.stack), p == pend ? REG_EBRACE : REG_BADBR));
| ~ ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
pre.i:8:74: error: expected ';' after return statement
8 | return (free (compile_stack.stack), p == pend ? REG_EBRACE : REG_BADBR));
| ^
| ;
1 warning and 1 error generated.
While there actually is an implicit conversion happening in C, it's not happening in C++. The warning is emitted for cpp compatibility, so it should not be emitted in this case.
A fix was introduced for the case:
void free ();
typedef enum {REG_EESCAPE} reg_errcode_t;
typedef struct {int *stack;} compile_stack_type;
reg_errcode_t byte_regex_compile () {
compile_stack_type compile_stack;
return (free (compile_stack.stack), REG_EESCAPE);
}
But we probably need a more generic fix maybe ?