在 C 中通过引用传递结构 [英] Pass struct by reference in C

问题描述

这个代码正确吗?它按预期运行，但是这段代码是否正确使用了结构的指针和点表示法?

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

struct someStruct {
 unsigned int total;
};

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

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

推荐答案

您对指针和点表示法的使用很好.如果有问题，编译器应该给你错误和/或警告.

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

这是您的代码的副本，其中包含一些额外的注意事项以及就结构体、指针和函数的使用以及变量的作用域而言需要考虑的事项.

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.

注意: 下面源代码示例中的代码编写差异是我在函数定义/声明中的结构名称之后和星号之前放置了一个空格，如 struct someStruct *p1; 和 OP 在星号后放置一个空格，如 struct someStruct* p1;.编译器没有区别，只是程序员的可读性和习惯不同.我更喜欢将星号放在变量名旁边，以明确星号会更改它旁边的变量名.如果声明或定义中有多个变量，这一点尤其重要.编写 struct someStruct *p1, *p2, var1; 将创建两个指针，p1 和 p2，以及一个变量，var1.编写 struct someStruct* p1, p2, var1; 将创建单个指针，p1 和两个变量 p2 和 var1

Note: A code writing difference in the source example below is I put a space after the struct name and before the asterisk in the function definition/declaration as in struct someStruct *p1; and the OP put a space after the asterisk as in struct someStruct* p1;. There is no difference to the compiler, just a readability and habit difference for the programmer. I prefer putting the asterisk next to the variable name to make clear the asterisk changes the variable name it is next to. This is especially important if I have more than one variable in a declaration or definition. Writing struct someStruct *p1, *p2, var1; will create two pointers, p1 and p2, and a variable, var1. Writing struct someStruct* p1, p2, var1; will create single pointer, p1 and two variables p2 and var1

// 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
", 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
", 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
", s.total);

    return 0;
}

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