通过函数传递结构时收到警告

Question

I'm trying to pass a whole structure through a function by value, and I test if the pass was successful by displaying the variables in the structure that was passed.

When I compile and run the program, the variables display correctly, but I get a warning:

warning: parameter names (without types) in function declaration

So it refers to the function prototype being declared with parameter names that had no types? But if I declared the function correctly, why am I getting this warning?

Also, in the short program I wrote to test this, the warning doesn't affect the fact that the output being displayed is correct.

But if I were to get this warning under the same circumstances (passing struct through function) when writing a larger-scale program, will it affect the output in that program being correct?

My code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void pass(Info);

typedef struct {
    char str[10];
    int num;
} Info;

void main() {
    Info i;

    strcpy(i.str, "Hello!");
    i.num = 1;

    pass(i);
}

void pass(Info i) {
    printf("%s\n", i.str);
    printf("%d\n", i.num);
}

Output:

Hello!

1

Answer 1

You have the typedef statement after the usage of the newly defined type ( synonym for another type, to be nitpicky ). At that point, in the function forward declaration, compiler does not know a type named Info . So, it assumes it to be a variable name.

Move the typedef before function prototype.

That said, void main() is not a valid signature for main() in a hosted environment. It should be int main(void) , at least.

Answer 2

Because you are declare the function prototype before the Info structure declaration.

An alternative could be the following code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct 
{
    char str[10];
    int num;
} Info;

void pass(Info);

void main() 
{
      Info i;

      strcpy(i.str, "Hello!");
      i.num = 1;
      pass(i)
}

void pass(Info i) 
{
     printf("%s\n", i.str);
     printf("%d\n", i.num);
}

Answer 3

The compiler considers this function declaration

void pass(Info);

as a function declaration with an identifier list. That is here Info is identifier of the parameter not its type.

However according to the C Standard (6.7.6.3 Function declarators (including prototypes)):

3 An identifier list in a function declarator that is not part of a definition of that function shall be empty .

SO the compiler issues a warning that the identifier list is not empty for the function declaration that is not a function definition.

warning: parameter names (without types) in function declaration

Either you should write

void pass();

Or you should place the typedef of the structure declaration before the function declaration and in this case the name Info will denote a type.