CODEWITHSUNDEEP
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Yaser Jaradeh
Yaser Jaradeh

Reputation: 322

Pass By Reference in C using linux programming

Is this code legal in C? I'm getting an error for the & sign. I am using Eclipse C/C++ IDE for Ubuntu to make this process easier.

void is_done(int &flag , char* ptr)
{
    int i=0;
    for(i=0;i<3;i++)
    {
        if(*ptr[i][0]==*ptr[i][1]==*ptr[i][2]||*ptr[0][i]==*ptr[1][i]==*ptr[2][i])
        {
            flag=1;
            return;
        }
    }
    if(*ptr[0][0]==*ptr[1][1]==*ptr[2][2]||*ptr[0][2]==*ptr[1][1]==*ptr[2][0])
    {
        flag=1;
        return;
    }
}

GCC gives me an error:

expected ‘;’, ‘,’ or ‘)’ before ‘&’ token
ipttt.c /OS line 7  C/C++ Problem

I really don't understand this error.

Upvotes: 1

Views: 723

Answers (4)

Mike
Mike

Reputation: 49403

What you wanted was to pass a pointer to an int like this:

void is_done(int *flag , char* ptr)
{
            // Then you must deference the variable to set the value
    *flag = 1; // or whatever value you want

Then you'd call your function with the flag like this:

int main()
{
   int flag = 0;
   char * ptr = NULL;
   ...
   is_done(&flag, ptr);  // Note that's not "reference" here, that's the address of
                         // your local flag variable

Of course you could just use a pointer, but since you were trying to "pass by reference" I assume you were not using a pointer in your code to begin with.

Upvotes: 1

Some programmer dude
Some programmer dude

Reputation: 409176

References using the & character in declarations is a C++ thing. To pass "objects" as a reference in C you have to use pointers:

void is_done(int *flag , char* ptr)
{
    ...

    *flag = 1;

    ...
}

Upvotes: 1

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726599

There is no "pass by reference" in C: that's C++. The only option available in C to accomplish this is passing by pointer

void is_done(int *flag , char* ptr)
{
    ...
    *flag=1;
    ...
}

You also need && with these chains of ==: they compile, but they do not do what you want them to do:

// DOES NOT WORK !!!
if(*ptr[0][0]==*ptr[1][1]==*ptr[2][2]||*ptr[0][2]==*ptr[1][1]==*ptr[2][0])

You need this:

if((*ptr[0][0]==*ptr[1][1] && *ptr[0][0]==*ptr[2][2]) || (*ptr[0][2]==*ptr[1][1] && *ptr[0][2]==*ptr[2][0])) {
    ...
}

Upvotes: 2

Axel
Axel

Reputation: 14159

C doesn't have references. Your code is C++. In C, you have to use pointers:

void is_done(int *flag , char* ptr)
{
    ...
    *flag = 1;
    ...
}

Upvotes: 4

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