将结构变量传递给函数

Question

I'm studying on structures in C programming. But, I'm confused in this code so that I don't understand. From where is the b coming in the function ? How can be a structure used like this ? Could you explain me ? Can we say display(struct book b1) ; calling the function ? Thank you for all appreciated answers.

#include <stdio.h>

struct book
{
    char name[25] ;
    char author[25] ;
    int callno ;
} ;
int main()
{
    struct book b1 = { "Let us C", "YPK", 101 } ;
    display ( b1 ) ;

    return 0;
}

void display ( struct book b )
{
    printf ( "\n%s %s %d", b.name, b.author, b.callno ) ;
}

Answer 1

b is the name of the parameter for the display function. It is what you have to pass to it. So in the main function when you call display(b1); b in the display function represents the book structure defined by b1 in the main function.

Answer 2

My best guess is that you are confused by the similarity of the variable name in main b1 , and the parameter name in the function b . Those names are completely unrelated, and could be called anything you like.

In the main function, b1 is a local variable that is declared as type struct book , and then initialized with a compile time constant initializer. The name b1 is arbitrary and can be any valid identifier.

In the display function, b is an argument to the function of type struct book . When the function is called, the caller must provide a struct book , and that struct will be copied into b . It's important to understand that b is a copy of the struct that was passed to the display function, which means that any changes that b makes to its local copy will not be propagated to the original structure declared in main .

Here's an attempt to demonstrate these principles in code

#include <stdio.h>

struct book
{
    char name[25] ;
    char author[25] ;
    int callno ;
};

void display( struct book someArbitraryNameForTheLocalCopyOfTheStructThatThisFunctionUses )
{
    printf ( "%s  ", someArbitraryNameForTheLocalCopyOfTheStructThatThisFunctionUses.name   );
    printf ( "%s  ", someArbitraryNameForTheLocalCopyOfTheStructThatThisFunctionUses.author );
    printf ( "%d\n", someArbitraryNameForTheLocalCopyOfTheStructThatThisFunctionUses.callno );

    // the following line has no effect on the variable in main since
    // all we have here is a local copy of the structure
    someArbitraryNameForTheLocalCopyOfTheStructThatThisFunctionUses.callno = 5555;
}

int main()
{
    struct book someLocalVariable = { "Let us C", "YPK", 101 };

    // the following line will make a copy of the struct for the 'display' function
    display( someLocalVariable );

    // the following line will still print 101, since the 'display' function only changed the copy
    printf ( "%d\n", someLocalVariable.callno );

    struct book anotherBook = { "Destiny", "user3386109", 42 };

    // any variable of type 'struct book' can be passed to the display function
    display( anotherBook );

    return 0;
}

Answer 3

Passing structures works like passing any other type: The function expects a variable of type struct b as it's argument, and then it just works with it. What is happening behind the scenes is, that all the data of b1 in your main function is copied into b of your display function. So be careful about that: When you change the value of a member of b in display , it won't change the value of b1 in main . If you want that to happen, you have to pass a pointer.