在函数内部更改堆栈上的结构

Question

我正在学习 艰难地学习 C ，我目前卡在 extra credit exercise number 16。

我目前正在尝试调整他们的代码并在堆栈而不是堆上创建结构，但我的代码出现分段错误，我不确定为什么或如何继续。非常感谢任何建议。

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

struct Person {
    char *name;
    int age;
    int height;
    int weight;
};

void Person_create(struct Person p,char *name,int age,int height,int weight)
{
    p.name = name;
    p.age = age;
    p.height = height;
    p.weight = weight;
}

void Person_print(struct Person who)
{
    printf("Name: %s\n", who.name);
    printf("\tAge: %d\n", who.age);
    printf("\tHeight: %d\n", who.height);
    printf("\tWeight: %d\n", who.weight);
}

int main(int argc, char *argv[])
{
    // make two people structures
    struct Person joe;
    struct Person frank;
    Person_create(
            joe,"Joe Alex",32,64,140);

    Person_create(
            frank,"Frank Blank",20,72,180);

    // print them out
    Person_print(joe);
    Person_print(frank);

    // make everyone age 20 years and print them again
    joe.age += 20;
    joe.height -= 2;
    joe.weight += 40;
    Person_print(joe);

    frank.age += 20;
    frank.weight += 20;
    Person_print(frank);

    return 0;
}

Answer 1

您的代码在运行时产生错误的原因是 C 按值传递 structs，这意味着 Person_create 中的以下赋值无效：

p.name = name;
p.age = age;
p.height = height;
p.weight = weight;

这不会更改 main() 内的 joe 和 frank，导致打印时出现未定义的行为，因为 name 数据成员 struct Person保持未初始化状态。

为了解决这个问题，通过指针传递 struct Person，并应用 -> 运算符代替 . 运算符：

void Person_create(struct Person *p,char *name,int age,int height,int weight)
{
    p->name = name;
    p->age = age;
    p->height = height;
    p->weight = weight;
}
...
Person_create(&joe, "Joe Alex", 32, 64, 140);
//            ^ Pass an address

It is asking for me to not use pointers or malloc, and says as advice to look into creating a struct on the stack

您的代码已经在堆栈上创建了 struct。如果您想完全摆脱指针，请将 Person_create 更改为 *return` 一个新的人，像这样：

struct Person Person_create(char *name,int age,int height,int weight) {
    struct Person p;
    p.name = name;
    p.age = age;
    p.height = height;
    p.weight = weight;
    return p;
}
...
joe = Person_create("Joe Alex", 32, 64, 140);

Answer 2

按以下方式定义函数Person_create

void Person_create(struct Person *p, char *name, int age, int height, int weight)
{
    p->name = name;
    p->age = age;
    p->height = height;
    p->weight = weight;
}

并这样称呼它

Person_create( &joe, "Joe Alex", 32, 64, 140 );

更复杂的方法如下

#include <string.h>
#include <stdlib.h>

//...

void Person_create(struct Person *p, const char *name, int age, int height, int weight)
{
    size_t n = strlen( name );
    p->name = malloc( ( n + 1 ) * sizeof( char ) );
    strcpy( p->name, name );
    p->age = age;
    p->height = height;
    p->weight = weight;
}

但是在这种情况下，您应该记得在结构对象离开其作用域时释放分配的内存。

Answer 3

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

typedef struct {
        char *name;
        int age;
        int height;
        int weight;
}Person;

void Person_print(Person who)
{
        printf("Name: %s\n", who.name);
        printf("\tAge: %d\n", who.age);
        printf("\tHeight: %d\n", who.height);
        printf("\tWeight: %d\n", who.weight);
}

int main(int argc, char *argv[])
{
        Person joe= {"Joe Alex", 32, 64, 140};
        Person frank={"Joe Alex", 32, 64, 140};

        printf("Joe is at memory location: %p\n", &joe);
        Person_print(joe);

        printf("Frank is at memory location: %p\n", &frank);
        Person_print(frank);

        joe.age += 20;
        joe.height -= 2;
        joe.weight += 40;
        Person_print(joe);

        frank.age += 20;
        frank.weight += 20;
        Person_print(frank);

        return 0;
}

typedef struct called 'Person', 那么你就不必使用 'Person create'.

在函数内部更改堆栈上的结构

Changing a struct on the stack inside function

c

stack

struct