在C中通过引用传递struct

Is this code correct? It runs as expected, but is this code correctly using the pointers and dot notation for the struct?

这段代码是否正确？它按预期运行，但这个代码是否正确使用结构的指针和点表示法？

struct someStruct {
 unsigned int total;
};

int test(struct someStruct* state) {
 state->total = 4;
}

int main () {
 struct someStruct s;
 s.total = 5;
 test(&s);
 printf("\ns.total = %d\n", s.total);
}

5 个解决方案

#1

Your use of pointer and dot notation is good. The compiler should give you errors and/or warnings if there was a problem.

你使用指针和点符号是好的。如果出现问题，编译器应该给出错误和/或警告。

Here is a copy of your code with some additional notes and things to think about so far as the use of structs and pointers and functions and scope of variables.

下面是代码的副本，其中包含一些额外的注释和事项，可以考虑使用结构和指针以及函数和变量范围。

// Define the new variable type which is a struct.
// This definition must be visible to any function that is accessing the
// members of a variable of this type.
struct someStruct {
    unsigned int total;
};

/*
 * Modifies the struct that exists in the calling function.
 *   Function test() takes a pointer to a struct someStruct variable
 *   so that any modifications to the variable made in the function test()
 *   will be to the variable pointed to.
 *   A pointer contains the address of a variable and is not the variable iteself.
 *   This allows the function test() to modify the variable provided by the
 *   caller of test() rather than a local copy.
 */
int test(struct someStruct *state) {
    state->total = 4;
    return 0;
}

/* 
 * Modifies the local copy of the struct, the original
 * in the calling function is not modified.
 * The C compiler will make a copy of the variable provided by the
 * caller of function test2() and so any changes that test2() makes
 * to the argument will be discarded since test2() is working with a
 * copy of the caller's variable and not the actual variable.
 */
int test2(struct someStruct state) {
    state.total = 8;
    return 0;
}

int test3(struct someStruct *state) {
    struct someStruct  stateCopy;
    stateCopy = *state;    // make a local copy of the struct
    stateCopy.total = 12;  // modify the local copy of the struct
    *state = stateCopy;    /* assign the local copy back to the original in the
                              calling function. Assigning by dereferencing pointer. */
    return 0;
}

int main () {
    struct someStruct s;

    /* Set the value then call a function that will change the value. */
    s.total = 5;
    test(&s);
    printf("after test(): s.total = %d\n", s.total);

    /*
     * Set the value then call a function that will change its local copy 
     * but not this one.
     */
    s.total = 5;
    test2(s);
    printf("after test2(): s.total = %d\n", s.total);

    /* 
     * Call a function that will make a copy, change the copy,
       then put the copy into this one.
     */
    test3(&s);
    printf("after test3(): s.total = %d\n", s.total);

    return 0;
}

#2

That's correct usage of the struct. There are questions about your return values.

这是结构的正确用法。您的退货值存在疑问。

Also, because you are printfing a unsigned int, you should use %u instead of %d.

另外，因为您正在打印unsigned int，所以应该使用％u而不是％d。

#3

Yes, that's right. It makes a struct s, sets its total to 5, passes a pointer to it to a function that uses the pointer to set the total to 4, then prints it out. -> is for members of pointers to structs and . is for members of structs. Just like you used them.

恩，那就对了。它创建一个struct，将其总数设置为5，将指向它的指针传递给一个函数，该函数使用指针将总数设置为4，然后将其打印出来。 - >是指向结构和指针的成员。是结构的成员。就像你使用它们一样。

The return values are different though. test should probably be void, and main needs a return 0 at its end.

但返回值不同。测试应该是无效的，并且main需要在其结尾处返回0。

#4

Yep. It's correct. If it wasn't (from the . / -> point of view), your compiler would yell.

是的。这是正确的。如果不是（从./->的角度来看），你的编译器会大喊大叫。

#5

Yes, its correct usage of structures. You can also use

是的，它正确使用结构。你也可以使用

typedef struct someStruct {
 unsigned int total;
} someStruct;

Then you won't have to write struct someStruct s; again and again but can use someStruct s; then.

那么你就不必编写struct someStruct了;一次又一次，但可以使用someStruct s;然后。

#1