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Is there any difference in below two castings ?
int a=10;
int *p=&a;
(void)p; //does not give any warning or error
or
(void *)p; //error: statement with no effect [-Werror=unused-value]
when complied with gcc -Wall -Werror --std=c99 -pedantic
Just saw that in this answer. (clearly I misunderstood something )
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Any pointer to an object, optionally type-qualified, can be converted to void* , keeping the same const or volatile qualifications.
Casting to void is used to suppress compiler warnings. The Standard says in §5.2. 9/4 says, Any expression can be explicitly converted to type “cv void.” The expression value is discarded.
You can cast a pointer to another pointer of the same IBM® i pointer type. Note: If the ILE C compiler detects a type mismatch in an expression, a compile time error occurs. An open (void) pointer can hold a pointer of any type.
The pointer is passed by value, so in f1 you're only changing a copy of the address held by a . If you want to change the pointer, i.e. allocate new memory for the pointer passed in, then you'll need to pass a pointer to a pointer: void f1(void **a) { // ... *a = malloc(sizeof(int)); // ...
When you do
(void) p;
You tell the compiler to simply disregard the result of the expression p. It's effectively the same as an empty statement:
;
When you do
(void *) p;
You tell the compiler to treat the variable p as a generic pointer, and that's the full expression for the statement, an expression which doesn't really do anything and you get the error message.
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