Struct and pointer

[rassul]

Hello, please someone show me the code for what is asked the in 
the following program's comments for the variable ptr.

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

struct _TEST
{
    char msg[250];
    int cnt;
}TEST;

typedef struct _TEST **PTR;

int main(int argc, char *argv[])
{
    PTR ptr;
    
    // allocate memory
    
    
    // assign values to the struct members
    
    
    // print the values
    
    
    // free the memory
    
    return 0;
}

[CommonTater]

Homework Policy

[_Mike]

So you're asking for someone to do your homework for you? Sure, I'll bite..

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

struct _TEST
{
    char msg[250];
    int cnt;
}TEST;

typedef struct _TEST **PTR;

int main(int argc, char *argv[])
{
    PTR ptr;
    
    // allocate memory
    malloc(1);
    
    // assign values to the struct members
    TEST.cnt = 0;
    memset(&TEST.msg, 0, sizeof(TEST.msg));
    
    // print the values
    printf("the values");
    
    // free the memory
    free(malloc(1));

    return 0;
}

Edit:
Damn, CommonTater was faster than me

[rassul]

Well, I am not a student rather an old programmer trying to refresh my C language. I actually have the following but it crashes when I ran it.

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

struct _TEST
{
    char msg[250];
    int cnt;
}TEST;

typedef struct _TEST **PTR;

int main(int argc, char *argv[])
{
    PTR ptr;
    
    // allocate memory
    if((ptr = (struct _TEST **) malloc(sizeof(struct _TEST))) == NULL) 
        return -1;
        
    // assign values to the struct members
    strcpy((*ptr)->msg, "This is what you think!");
    (*ptr)->cnt = 2011;
    
    // print the values
    printf("%s %d\n", (*ptr)->msg, (*ptr)->cnt);
    
    // free the memory
    free(ptr);
    
    return 0;
}

[quzah]

What's the point of doing this:

typedef struct _TEST **PTR;

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

If you are just doing this:

if((ptr = (struct _TEST **) malloc(sizeof(struct _TEST))) == NULL) 
        return -1;

You don't even need the cast, and you would have been better just doing:

ptr = malloc( sizeof **ptr );

Except of course that would still be wrong, because you have a struct foo ** and you are allocating size of struct foo instead of struct foo *.

FYI: No one likes typedefed pointers.


Quzah.

[_Mike]

You should have posted that code directly even though it doesn't work. Most people tend to look down on "write this for me" type of posts and assume the writer is to lazy to do it him/her self.

Is there any special reason you have to use a pointer to pointer to _TEST or is that a mistake?
When you try to dereference it here

strcpy((*ptr)->msg, "This is what you think!");

You first dereference ptr which points to a struct _TEST, but because you declared it as _TEST** the compiler thinks it points to a pointer to _TEST, which makes the result an invalid pointer.
You then dereference the invalid pointer with -> which causes the crash.

[rassul]

This is not an assignment nor a project, simply reviewing struct, typedef, and pointers in combination for learning purposes, because in real world you will this kind code a lot!

[rassul]

It crashes and does not even print "1- debug..."

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

struct _TEST
{
    char msg[250];
    int cnt;
}TEST;

typedef struct _TEST **PTR;

int main(int argc, char *argv[])
{
    PTR ptr;
    
    // allocate memory
    if((ptr = malloc(sizeof **ptr)) == NULL) 
        return -1;
    printf("1- debug...\n");
        
    if((*ptr = malloc(sizeof TEST)) == NULL)
        return -2;
        
    printf("2- debug...\n");
    // assign values to the struct members
    strcpy((*ptr)->msg, "This is what you think!");
    (*ptr)->cnt = 2011;
    
    // print the values
    printf("%s %d\n", (*ptr)->msg, (*ptr)->cnt);
    
    // free the memory
    free(ptr);
    
    return 0;
}

[quzah]

This is not an assignment nor a project, simply reviewing struct, typedef, and pointers in combination for learning purposes, because in real world you will this kind code a lot!

So are you saying the real world is full of broken code, or full of typedefed pointers, or both? The real world sounds pretty crappy.


Quzah.

[quzah]

typedef struct _TEST **PTR;

int main(int argc, char *argv[])
{
    PTR ptr;
    
    // allocate memory
    if((ptr = malloc(sizeof **ptr)) == NULL) 
        return -1;
    printf("1- debug...\n");

You are still doing that wrong. If you want a foo** returned, then allocate the size of a foo**, not the size of a foo.

struct foo **p = malloc( sizeof *p ); /* allocate one pointer to foo */
p[0] = malloc( sizeof **p ); /* allocate one actual structure */

There's a whole bunch of posts about dynamic 2D arrays around here which cover the types of allocations you are doing. You should probably find some and read them.


Quzah.

[_Mike]

It crashes and does not even print "1- debug..."

printf is buffered and therefore not very good for diagnosing crashes. Print to stderr or better yet use a debugger.

[AndrewHunter]

So are you saying the real world is full of broken code, or full of typedefed pointers, or both? The real world sounds pretty crappy.
Quzah.

Quzah, what do you know of the real world? This guy obviously knows what he is talking about, after all he is an 'old programmer trying to refresh on c'. I mean you don't have any real experience. You and Salem are just some geeky teenagers, right?

[quzah]

You and Salem are just some geeky teenagers, right?

Yes. I started posting here when I was three.


Quzah.

[Matticus]

Yes. I started posting here when I was three.

Quzah.

It shows.

(Here is where the typical person would place a "winked smiley" to show humor. Unfortunately, I'm a curmudgeon who doesn't deal well with that crap. I'll hope the the jest on the sassy response of quzah is understood for what is it - a poor, albeit earnest, humorous response in the language it was received in. And I don't mean 'C'.)

[quzah]

And I don't mean 'C'.)

Then how am I supposed to understand it?


Quzah.