结构成员char *在函数调用中初始化为参数

Question

这更多是出于好奇，而不是我想要或需要它的方式。 但是，当我做一些样机并想测试某些东西时，我最终得到了这样的东西……并且想知道为什么它不能按我预期的那样工作。

typedef struct {
    char *a;
    char *b;
    char *c;
}mystruct;

void init_chars (char *arg)
{
    arg = malloc (sizeof (char)*10);
    arg = "0123456789";
    printf ("%s\n", arg);
}

int main ()
{
    mystruct *msp = malloc (sizeof (mystruct));
    init_chars (msp->a);
    init_chars (msp->b);
    init_chars (msp->c);
    printf ("%s, %s, %s\n", msp->a, msp->b, msp->c);
    return 0;
}

打印...

0123456789

0123456789

0123456789

(null),(null),(null)

Answer 1

将值传递给函数参数时，C中有两件事

1. 传递价值
2. 通过参考

您正在按值传递，因此您要将一些未初始化的值传递给该函数，并在该函数外部看不见的函数中对其进行初始化，因此您试图在main()使用未初始化的值，这将导致未定义的行为。

arg = "0123456789";

如果要将字符串复制到某个内存位置，则需要memcpy()或strcpy()

注意：使用未初始化的值会导致未定义的行为。

Answer 2

您的代码有几个问题。 这是固定代码：

/* Don't forget to include <stdio.h>, <stdlib.h> and <string.h> */

typedef struct {
    char *a;
    char *b;
    char *c;
}mystruct;

void init_chars (char **arg) /* This should be char** as the address of a char* is passed */
{
    *arg = malloc ( /*sizeof(char)* */ 11); /* sizeof(char) is 1 always. You don't need it */
                                            /* Note the change of `arg` to `*arg` too */
                                            /* And that I've used 11 and not 10 because there needs to be space for the NUL-terminator */

    if(*arg == NULL) /* If the above malloc failed */
    {
        printf("Oops! malloc for *arg failed!");
        exit(-1);
    }

    //arg = "0123456789"; /* You just lost the allocated memory as you make the pointer point to a string literal */

    strcpy(*arg, "0123456789"); /* Use strcpy instead */
    printf ("%s\n", *arg);      /* *arg here */
}

int main ()
{
    mystruct *msp = malloc (sizeof (mystruct));

    if(msp == NULL) /* If the above malloc failed */
    {
        printf("Oops! malloc for msp failed!");
        return -1;
    }

    init_chars (&(msp->a));
    init_chars (&(msp->b)); 
    init_chars (&(msp->c));  /* Pass address of variables rather than their value */

    printf ("%s, %s, %s\n", msp->a, msp->b, msp->c);

    free(msp->a);
    free(msp->b);
    free(msp->c);

    free(msp);    /* Free everything after use */

    return 0;
}

Answer 3

typedef struct {
    char *a;
    char *b;
    char *c;
}mystruct;

void init_chars (char **arg)
{
    *arg = malloc (sizeof (char)*11);
    if(*arg == NULL)
    {
        printf("malloc failed!");
        exit(-1);
    }
    strcpy(*arg,"0123456789");
    printf ("%s\n", *arg);
}

int main ()
{
    mystruct *msp = malloc (sizeof (mystruct));
    if(msp == NULL)
    {
        printf("malloc failed");
        exit(-1);
    }
    init_chars (&msp->a);
    init_chars (&msp->b);
    init_chars (&msp->c);
    printf ("%s, %s, %s\n", msp->a, msp->b, msp->c);
    free(msp->a);
    free(msp->b);
    free(msp->c);
    free(msp)
    return 0;
}

输出：

0123456789

0123456789

0123456789

0123456789、0123456789、0123456789

与您的代码进行比较，您将理解为什么