Memory Allocation Using String (Please Help)

Code:

---

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#define size 10
void main()
{
        int i,s;
        char **names;
        printf("Enter Size Of Array..\n");
        scanf("%d",&s);
        fflush(stdin);
        names=(char **)malloc(s*sizeof(char));
        for(i=0;i<s;i++)
        {
                names[i]=(char *)malloc(size*sizeof(char));
        }
        printf("\nEnter Data..\n");
        for(i=0;i<s;i++)
        {
                scanf("%s",&names[i][0]);
        }
        printf("\nDisplaying Data..\n");
        for(i=0;i<s;i++)
        {
                printf("%s",names[i][0]);
        }
        free(names);
        getch();
}

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I Don't Know What Seems To Be The Problem, What I'm trying to do is to read a set of names using memory allocation and print them and this doesn't work!

Help Would Be Really Appreciated.

Code:

---

#include<stdio.h>
#include<conio.h> //delete this
#include<stdlib.h>
#define size 10
void main() //see faq void main
{
    int i,s;
    char **names;
    printf("Enter Size Of Array..\n");
    scanf("%d",&s);
    fflush(stdin); //see faq fflushing stdin
    names=(char **)malloc(s*sizeof(char)); //see faq casting malloc
    for(i=0;i<s;i++)
    {
        names[i]=(char *)malloc(size*sizeof(char)); //you don't free these
    }
    printf("\nEnter Data..\n");
    for(i=0;i<s;i++)
    {
        scanf("%s",&names[i][0]);
    }
    printf("\nDisplaying Data..\n");
    for(i=0;i<s;i++)
    {
        printf("%s",names[i][0]); //this is a character, not a string. you want names[i]
    }
    free(names);
    getch(); //use getchar. it's standard, getch is not.
}

---

Code:

---

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#define size 10
void main()
{
        int i,s;
        char **names;
        printf("Enter Size Of Array..\n");
        scanf("%d",&s);
        fflush(stdin);
        names=(char **)malloc(s*sizeof(char));
        for(i=0;i<s;i++)
        {
                names[i]=(char *)malloc(size*sizeof(char));
        }
        printf("\nEnter Data..\n");
        for(i=0;i<s;i++)
        {
                scanf("%s",names[i]);
        }
        printf("\nDisplaying Data..\n");
        for(i=0;i<s;i++)
        {
                printf("%s",names[i]);
        }
        free(names);
        getch();
}

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There is nothing wrong with pointer casting. Your problem is about printf, scanf string implementation. Examine and train it.

> There is nothing wrong with pointer casting.
Only that it hides mistakes which the compiler might otherwise be able to tell you about.
For example, hiding the fact that you didn't prototype malloc properly because you didn't include stdlib.h (for example), which is also mentioned in the FAQ.

Another problem:

Code:

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names=(char **)malloc(s*sizeof(char));

---

You are allocating the size of chars (1 byte), not a pointer to char (typically 4 bytes).

And to top it off, you are only freeing the pointer names, and not the pointers in the array names.
Plus you are reading your strings in an extremely unsafe way. You shouldn't. You have some reading to do:
http://cpwiki.sourceforge.net/buffer_overrun

Quote:

---

Originally Posted by Salem

> There is nothing wrong with pointer casting.
Only that it hides mistakes which the compiler might otherwise be able to tell you about.
For example, hiding the fact that you didn't prototype malloc properly because you didn't include stdlib.h (for example), which is also mentioned in the FAQ.

---

And on certain compilers/processor architectures, this could have extreme consequences. In Motorola 68K processor family, registers are either "data" or "address" registers. Pointers are returned in address registers, integer return values are in data registers. An undefined function will return an integer, so if you don't include stdlib.h, the compiler will assume that you get an integer (that is later cast to an pointer) - so it will use the wrong register, and random bad things will happen [it is quite possible that the malloc function itself is temporarily doing math on the allocated memory, thus keeping something akin to the allocated location in register D0 which is used by the compiler to return an integer value, so it may not even crash on the initial instance!].

Similarly, in an x86-64 architecture, the return value type of integer is 32-bit. A pointer is 64-bit, and the consequence would be "lost the upper 32-bits".

--
Mats