为什么不能在函数中修改结构的成员变量？

Question

I am curious to why one cannot modify the variables of a struct when passed as a parameter into a function. I understand that the parameter is pass by value but when passing a struct variable you are passing its reference as the value.

Surely C doesn't create a copy of the struct on the stack, because the value of the struct reference being passed in is the same as the value of passing a pointer of the struct into a function.

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


typedef struct String {
    char* name;
    int x;
} String;

/* Func Protos */
void foo(String str);
void bar(String * str);
void banjo(String str);

int main() {
    String str = *(String *)malloc(sizeof(String));
    str.x = 5; /* x == 5 */
    foo(str); /* x == 5 still */
    printf("%d\n", str.x);
    banjo(str); /* x == 5 still */
    printf("%d\n", str.x);
    bar(&str); /* and of course, x = 0 now */
    printf("%d\n", str.x);
}

void foo(String str) {
    printf("%d\n", str); /* Same value as if we passed String * str */
    str.x = 0; 
}

void bar(String * str) {
    printf("%d\n", *str);
    str->x = 0;
}

void banjo(String str) {
    printf("%d\n", str);
    String * strptr = &str; /* This should be identical to passing String * str */
    strptr->x = 0;
}

Produces this output:

3415000
5
3415000
5
3415000
0

Any help would be much appreciated!

Answer 1

void banjo(String str) {
    printf("%d\n", str);
    String * strptr = &str; /* This should be identical to passing String * str */
    strptr->x = 0;
}

C is pass-by-value. What you get in banjo function with str argument is a copy of main str object.

So your banjo function is equivalent to:

void banjo(String str) {
    printf("%d\n", str);
    strptr.x = 0;  // useless, the object is discarded after function returns
}

By the way printf("%d\\n", str); is not valid. d conversion specifier required an int but you are passing a structure value. The call invokes undefined behavior. If you want to print the address of str object use:

printf("%p\n", (void *p) &str);

Answer 2

This line:

String str = *(String *)malloc(sizeof(String));

Is why you're getting that behavior. Change it to:

String *str = (String *)malloc(sizeof(String));

And that will solve the problem.

Answer 3

The statement

printf("%d\n", str);

doesnot print the address of the structure, instead it prints the value of the first element
inside the structure. And since the first element inside your structure is a pointer to char, thus the printf
is giving garbage value on the console.

Try this example :

#include<stdio.h>
struct a
{
   int b;
};

int main()
{
     struct a a1;
     a1.b = 10;  // try giving different value to a1.b
     printf("b = %u\n", a1.b);
     printf("a1 = %u\n", a1);
    return 0;
}

Where you think you are printing the base address of the structure, you are actually
printing the value of the first element inside the structure.