如果传入的函数有参数，如何屏蔽来自 makecontext 的警告？

Question

在 C 语言 <ucontext_t.h> 中使用函数 makecontext 时，它采用参数 (ucontext_t* context, void (*someFunction)(void), int numberOfArguments, ...)。 numberOfArguments是传参给someFunction，但是传参时编译报错

我需要或者应该删除这些，因为这是针对学校项目的，警告可能会让我失分。我想我可以编辑 Makefile 的标志，但我更愿意调整代码。

我做了 2 次测试。

1: 传递了一个函数和字符串作为参数

test1 的代码是：

#include <stdio.h>
#include <ucontext.h>

// Outputs arg as a string
void* f(void* arg) {
    printf("%s\n", (char*)arg);
    return 0; // Returns back to c0
}

// makecontext takes as argument2, (void (*)(void)), but allows (void (*)(...))
// I want to remove the warning from using (void (*)(...))
void helper(void* (*selectFunction)(void*), void* selectArg) {
    // Calls select function with select argument
    selectFunction(selectArg);
}

int main()
{
    // Initialize contexts
    ucontext_t c0, c1;
    char stack[1024];

    getcontext(&c1);
    // Setup stack for c1
    c1.uc_stack.ss_sp = stack;
    c1.uc_stack.ss_size = sizeof(stack);
    c1.uc_link = &c0; // Swap to c0 stack on c1 stack end

    void* (*selectFunction)(void*) = &f;
    void* selectArg = (void*)("abc123 testing testing");

    // Passes function f with arg "abc123 testing testing" to helper
    makecontext(&c1, helper, 2, selectFunction, selectArg);

    // Swap to context that calls function helper
    swapcontext(&c0, &c1);

    printf("Welcome back\n");

    return 0;
}

我对 test1 的输出是：

$gcc -o main *.c
main.c: In function ‘main’:
main.c:33:22: warning: passing argument 2 of ‘makecontext’ from incompatible pointer type [-Wincompatible-pointer-types]
     makecontext(&c1, helper, 2, selectFunction, selectArg);
                      ^~~~~~
In file included from main.c:2:0:
/usr/include/ucontext.h:47:13: note: expected ‘void (*)(void)’ but argument is of type ‘void (*)(void * (*)(void *), void *)’
 extern void makecontext (ucontext_t *__ucp, void (*__func) (void),
             ^~~~~~~~~~~
$main
abc123 testing testing
Welcome back

2：传递了一个整数作为参数

test2 的代码是：

#include <stdio.h>
#include <ucontext.h>

// Outputs arg as a string
void f(int n) {
    printf("%d\n", n);
    return; // Returns back to c0
}

// makecontext takes as argument2, (void (*)(void)), but allows (void (*)(...))
// I want to remove the warning from using (void (*)(...))
void helper(int arg) {
    // Calls f with arg
    f(arg);
}

int main()
{
    // Initialize contexts
    ucontext_t c0, c1;
    char stack[1024];

    getcontext(&c1);
    // Setup stack for c1
    c1.uc_stack.ss_sp = stack;
    c1.uc_stack.ss_size = sizeof(stack);
    c1.uc_link = &c0; // Swap to c0 stack on c1 stack end

    // Make a context to run f with argument 5
    makecontext(&c1, helper, 1, 5);

    // Swap to context that calls function helper
    swapcontext(&c0, &c1);

    printf("Welcome back\n");

    return 0;
}

我对 test2 的输出是：

$gcc -o main *.c
main.c: In function ‘main’:
main.c:30:22: warning: passing argument 2 of ‘makecontext’ from incompatible pointer type [-Wincompatible-pointer-types]
     makecontext(&c1, helper, 1, 5);
                      ^~~~~~
In file included from main.c:2:0:
/usr/include/ucontext.h:47:13: note: expected ‘void (*)(void)’ but argument is of type ‘void (*)(int)’
 extern void makecontext (ucontext_t *__ucp, void (*__func) (void),
             ^~~~~~~~~~~
$main
5
Welcome back

Answer 1

参数someFunction的类型为void(*)(void)，需要传递一个指向不带参数的函数的指针和returns void（不是void *)

您可以安排它工作，将 someFunction 的定义更改为如下内容：

void *(*someFunction)()  /* don't specify the argument list */

在这种情况下，编译器将接受任何 returns void * 的函数，但是您必须仔细检查您传递的函数是否具有正确的参数数量和类型。这被称为不完整的类型定义，它很容易出错（它是从古老的 C 中保存下来的，在 ANSI 修订版之前很久）

如果您想要一个接受 void * 参数和 returns 一个 void * 结果的函数，请将 someFunction 的定义更改为：

void *(*someFunction)(void *), ...

但如果这是一个学术练习，我建议你做完整的类型检查，不要试图使用 C 工具来犯错误:)（他们要求你非常清楚你在做什么）

在你的例子中，辅助函数需要一个 int 参数，它不是 void * （实际上，它们很可能大小不一样）所以 makecontext 函数将无法正确调用它。编译器在那里保护你免受你自己的伤害。

Answer 2

这实际上是makecontext()的这个manual指出的一个错误：

With the incorporation of the ISO/IEC 9899:1999 standard into this specification it was found that the ISO C standard (Subclause 6.11.6) specifies that the use of function declarators with empty parentheses is an obsolescent feature. Therefore, using the function prototype:

void makecontext(ucontext_t *ucp, void (*func)(), int argc, ...);

is making use of an obsolescent feature of the ISO C standard. Therefore, a strictly conforming POSIX application cannot use this form. Therefore, use of getcontext(), makecontext(), and swapcontext() is marked obsolescent.

通常的解决方法是使用强制转换。为此，定义一个新类型：

typedef void (* ucfunc_t)(void);

然后你使用这个定义来转换传递给makecontext()的参数:

#include <stdio.h>
#include <ucontext.h>

typedef void (* ucfunc_t)(void);

// Outputs arg as a string
void* f(void* arg) {
    printf("%s\n", (char*)arg);
    return 0; // Returns back to c0
}

// makecontext takes as argument2, (void (*)(void)), but allows (void (*)(...))
// I want to remove the warning from using (void (*)(...))
void helper(void* (*selectFunction)(void*), void* selectArg) {
    // Calls select function with select argument
    selectFunction(selectArg);
}

int main()
{
    // Initialize contexts
    ucontext_t c0, c1;
    char stack[1024];

    getcontext(&c1);
    // Setup stack for c1
    c1.uc_stack.ss_sp = stack;
    c1.uc_stack.ss_size = sizeof(stack);
    c1.uc_link = &c0; // Swap to c0 stack on c1 stack end

    void* (*selectFunction)(void*) = &f;
    void* selectArg = (void*)("abc123 testing testing");

    // Passes function f with arg "abc123 testing testing" to helper
    makecontext(&c1, (ucfunc_t)helper, 2, selectFunction, selectArg);

    // Swap to context that calls function helper
    swapcontext(&c0, &c1);

    printf("Welcome back\n");

    return 0;
}

如果传入的函数有参数，如何屏蔽来自 makecontext 的警告？

How can I mask the warning from makecontext if the function passed in has parameters?

c

gcc

casting

ucontext