[英]Nested callback function in C
I have written some code for calling nested functions using callback.我已经编写了一些使用回调调用嵌套函数的代码。 But i am not getting output as I expected.
但我没有像预期的那样得到 output。
Please have look into code:请查看代码:
#include <stdio.h>
#include <stdlib.h>
typedef int (*f_ptr)(int a, int b);
typedef int (*pair_ptr)(f_ptr);
pair_ptr cons(int a, int b)
{
int pair(f_ptr f)
{
return (f)(a, b);
}
return pair;
}
int car(pair_ptr fun)
{
int f(int a, int b)
{
return a;
}
return fun(f);
}
int main()
{
int a = 3;
int b = 4;
// It should print value of 'a'
printf("%d", car(cons(a, b))); // Error : It is printing '0'
return 0;
}
I have also tried with function pointer but in that also i am getting the same output as above.我也尝试过使用 function 指针,但我也得到了与上面相同的 output。
Try this (maybe move functions and variables related to the closure to their own file):试试这个(也许将与闭包相关的函数和变量移动到它们自己的文件中):
#include <stdio.h>
typedef int (*f_ptr)(int a, int b);
typedef int (*pair_ptr)(f_ptr);
static int PairA, PairB;
static void setPairA(int a) { PairA = a; }
static void setPairB(int b) { PairB = b; }
int f(int a, int b) {
(void)b; // removed unused parameter warning
return a;
}
int pair(f_ptr fp) {
return fp(PairA, PairB);
}
pair_ptr cons(int a, int b) {
setPairA(a);
setPairB(b);
return pair;
}
int car(pair_ptr fun) {
return fun(f);
}
int main(void) {
int a = 3;
int b = 4;
printf("%d\n", car(cons(a, b)));
return 0;
}
Note that pair()
is not reentrant, nor can you call it with different values for PairA
and/or PairB
at the same time.请注意,
pair()
不可重入,您也不能同时为PairA
和/或PairB
使用不同的值来调用它。
C doesn't support nested function, but a GCC extension does. C 不支持嵌套 function,但 GCC 扩展支持。 The docs say:
文档说:
If you try to call the nested function through its address after the containing function exits, all hell breaks loose.
如果您尝试在包含 function 退出后通过其地址调用嵌套的 function,一切都会崩溃。
pair
tries to use a
and b
which no longer exist. pair
尝试使用不再存在a
和b
。 GCC might provide nested functions, but they don't provide closures . GCC 可能提供嵌套函数,但不提供闭包。 This means the values of
a
and b
aren't captured by the function;这意味着
a
和b
的值没有被 function 捕获; it's merely an address being returned by cons
.它只是
cons
返回的地址。
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