指针到指针数组中指针的访问地址

Question

4I am trying to get a pointer address from a pointer to a pointer array, however I just get garbage when I run the code below. How do I retrieve the pointer address of one of the char * from the char ** array? Thanks for checking this out.

Specifically, I want to obtain the address of "wh" via "what" and assign it to "hi".

char * hi;
char * wh;
char ** what;

int main(void) {
    char z[4] = "wx\0";
    char a[4] = "ab\0";
    hi = &z;
    wh = &a;
    what = (char **) malloc( 25 * sizeof(char));
    what[0] = &hi;
    what[1] = &wh;

    printf("%s | %s\n", hi, wh);

    hi = &what[1];

    printf("%s | %s\n", hi, wh);

    return EXIT_SUCCESS;
}

Answer 1

Corrected code:

#include <stdio.h>
#include<stdlib.h>
char * hi;
char * wh;
char ** what;

int main(void) {
    char z[3] = "wx\0";
    char a[3] = "ab\0";
    hi = z;
    wh = a;
    what = malloc(2 * sizeof *what);
    what[0] = hi;
    what[1] = wh;

    printf("%s | %s\n", hi, wh);

    hi = what[1];

    printf("%s | %s\n", hi, wh);

    return 0;
}

The type of hi and wh is not the same as of &z and &a , even though they should yield the same value. Also you want what[1] in hi not the value of the memory location where it is. Also, in C you do not need to cast the return of malloc. However, in C++ the cast is required although its better to use the new operator.

Answer 2

Pointer Assignment

• hi and wh are defined as pointer to char . So, change your assignment statements

  from,

   hi = &z; /* Type of &z is pointer-to-array-of-4-char */ wh = &a; /* Type of &a is pointer-to-array-of-4-char */ 

  to,

   hi = z; /* Type of z is pointer-to-char. Array name, here, decays into pointer to first element of the array */ wh = a; /* Type of a is pointer-to-char */ 

  ---

• what is defined as type, pointer to pointer to char . So, change your malloc statement

  from,

   what = (char **) malloc( 25 * sizeof(char)); /* what needs to hold char*'s, you have allocated only for char's */ 

  to,

   what = malloc(25 * sizeof *what); /* what is of type char**. *what gives you char* */ 

  ---

• Also, change your what , hi assignment statements

  from,

   what[0] = &hi; /* Type of what[0] is pointer-to-char; Type of &hi is pointer-to-pointer-to-char */ what[1] = &wh; /* Type of what[1] is pointer-to-char; Type of &wh is pointer-to-pointer-to-char */ ... hi = &what[1]; /* Type of hi is pointer-to-char; Type of &what[1] is pointer-to-pointer-to-char */ 

  to ,

   what[0] = hi; /* Types are same */ what[1] = wh; /* Types are same */ ... hi = what[1]; /* Types are same */ 

  ---

Note: The size of what is too high than what really required for the posted program. If you feel, you are going to have the same program, modify 25 to 2 in the malloc .

Answer 3

char * hi;
char * wh;
char ** what;

int main(void) {
    char z[4] = "wx\0";
    char a[4] = "ab\0";
    hi = &z;
    wh = &a;

First mistake; the types of the expressions &z and &a are both char (*)[4] (pointer to 4-element array of char ), not char * . To fix this, drop the & from both:

    hi = z;
    wh = a;

Except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize an array, an expression of type "N-element array of T " will be converted ("decay") to an expression of type "pointer to T ", and the value of the expression is the address of the first element in the array.

    what = (char **) malloc( 25 * sizeof(char));

Don't cast the result of malloc ; it isn't necessary ¹ , and depending on the compiler version it can suppress a useful diagnostic. You also have a type mismatch. You want to space for 25 pointers to char , but you're only allocating enough for 25 plain char s. Rewrite that as

    what = malloc( sizeof *what * 25 );

The type of the expression what is char ** ; thus, the type of the expression *what is char * . So sizeof *what will give you the same result as sizeof (char *) . The above line allocates enough memory to store 25 pointers to char , and assigns the resulting pointer to what .

    what[0] = &hi;
    what[1] = &wh;

Again, drop the & ; hi and wh are already pointers to char :

    what[0] = hi;
    what[1] = wh;

    printf("%s | %s\n", hi, wh);

    hi = &what[1];

Type mismatch; &what[1] has type char ** , hi has type char * . Again, drop the & .

    hi = what[1];

hi and wh now point to the same thing ( what[1] == wh and hi == what[1] , so hi == wh ; that is, both hi and wh contain the same pointer value).

    printf("%s | %s\n", hi, wh);

    return EXIT_SUCCESS;

}

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^{1. In C, that is; C++ is a different story, but if you're writing C++, you shouldn't be using malloc anyway.}