使用结构vs指向结构的指针

debugcn 发表于 Dev

大卫542

我以以下示例为例，以查看直接传递对象与传递指针之间的区别：

#include "stdio.h"

// Car object
typedef struct Car {
    char*           name;
    unsigned int    price;
} Car;

void print_car(Car car) {
    printf("<Car: %s, Price: $%d>", car.name, car.price);
};

void print_car2(Car *car) {
    printf("<Car: %s, Price: $%d>", car->name, car->price);
};

int main(int argc, char* argv[]) {

    Car chevy = {chevy.name = "Chevy", chevy.price = 45000};
    print_car(chevy);

    Car mazda = {chevy.name = "Mazda", chevy.price = 30000};
    print_car2(&mazda);

    return 1;

}

除了第一种方法对我而言更具可读性和易于理解之外，两者之间有什么区别？什么时候传递指针是唯一的选择，为什么在上述情况下两者都起作用？

通常（不仅对于结构体）将变量传递给函数会复制该变量，因此，如果您要更改此变量，则必须返回更改后的副本的值，但您可能希望更改变量并返回还有其他事情，在这种情况下，您别无选择，将指针作为参数exemple传递：

第一个例子是传递变量

int useless_func(int nb) /*nb is actually a copy of the variable passed as argument */
{
    nb++; /* the copy was incremented, not the real variable */
    return nb; /* the new value is returned */
}
int main()
{
    int nb = 1;

    useless_func(nb); /* here nb is still = 1 cause it wasn't altered by the functionalists */
    nb = useless_func(nb); /* here nb is = 2 cause it took the return value of the func */
}

现在是指针的第二个愚蠢的例子：

char *useless_func(int *nb) /* nb is now a pointer to the variable */
{
    *nb++; /* the derefencement of the pointer (so the variable value) was incremented */
    return strdup("This is a useless return"); /* we return some other stupid stuff */
}
int main()
{
    int nb = 1;
    char *str = useless_func(&nb); /* now nb is = 2 and str is an allocated useless string woohoo */
}

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