这些函数typedef是什么意思？

Question

I am trying to understand what the following typedefs mean. Are they function pointers?

typedef int Myfunc(char *);

static Myfunc myfunc;

int myfunc(char *string)
{
    printf("%s\n", string);
    return 0;
}

I know typedef int Myfunc(char *) means func Myfunc return an integer,that's all,all right? And I thought, next statement, how could call myfunc ? It should be this way static Myfunc *myfunc , mean a function pointer,isn't it?

Answer 1

The second line is a declaration of a function, not a function pointer. The function is of type MyFunc , is called myfunc , and has static linkage : meaning that the function is not available to other source files compiled into the same object.

Answer 2

The signature for the myfunc is: typedef int (*MyFunc)(char *); Then you can declare a variable of type MyFunc ie,

static MyFunc func_ptr;

You can then assign a function matching the signature to this variable.

Answer 3

调用myfunc与函数调用相同：

myfunc("a-string");

Answer 4

I'm not sure that's valid code.

typedef int (*Myfunc)(char *);

declares a type Myfunc , that is a pointer to a function that takes a char * and returns int .

You cannot forward declare a function with a typedef. Omit static Myfunc myfunc; and instead begin your function definition with

static int myfunc(char *string) {