结构成员char *在函数调用中初始化为参数

Question

This is more out of curiosity, and is not how I want or need it to work. But when I was doing some mockups and wanted to test something out, I ended up with something like this... and was wondering why it doesn't work as I expected.

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;
}

Prints...

0123456789

0123456789

0123456789

(null),(null),(null)

Answer 1

There are 2 things in C when it comes to passing the value to function parameters

1. Pass by value
2. Pass by reference

You are doing pass by value so you are passing some unintiialized value to the function and initializing it in the function which is not visible outside the function, you are trying to use uninitialized values in main() which will lead to undefined behavior.

arg = "0123456789";

If you want to copy a string to some memory location then you need memcpy() or strcpy()

Note: Using uninitialized values leads to undefined behavior.

Answer 2

Your code has several problems. Here is the fixed code:

/* 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;
}

output:

0123456789

0123456789

0123456789

0123456789, 0123456789, 0123456789

compare with your code u will understand why