c - 将结构的指针成员传递给函数时出现错误 )

Question

我是 C 的初学者。在我的程序中，我有一个结构和一个函数。我正在尝试传递结构中存在的指针作为函数中的参数。但它在点运算符处显示错误“Expected )”。这令人困惑，因为我的函数的其他参数也来自结构，但在这些结构中没有看到此错误。

我尝试将函数的返回类型更改为所有类型，但仍然没有。

struct signal
{
bool *input;
int previousop;
int n;
}s; //my structure
void noiseremove(bool *input, int n, int count1, int count0, bool 
previousop)//function i declared and defined before main function
{//my function here}
void main()
{
void noiseremove(bool *s.input , int s.n, int s.count1, int s.count0, bool 
s.previousop); //this is where i call the function and facing an error at 
*s.input 
}

我不确定我哪里出错了，或者语法是否有误。我希望该函数接受参数，但事实并非如此。

Answer 1

在另一个函数中包含函数是 not possible in C ...

因此，您的代码应该看起来像这样:

struct signal
{
    bool *input, previousop;
    int n, count0, count1;
} s;

void noiseremove(bool *input, int n, int count1, int count0, bool previousop)
{
    /* Try using multi-line comments since single-line comments can comment out the end
       braces as well...*/
}

void main()
{
    /* Initialize the structure before accessing any of its variables, or it will lead
       to undefined behavior! */
    s = {0};
    /* Don't declare the identifiers again... It is 'syntax error' and 
       your calling convention doesn't in the least look like a calling 
       convention but more like a function declaration with invalid identifiers
       for the parameters... */
    noiseremove(s.input , s.n, s.count1, s.count0, s.previousop);
}