strcat() and segmentation faults. But malloc() already done.

[Stanley S]

Hi,

I'm puzzled. Why does the following cause a seg fault? Notwithstanding
that I've already malloc() a certain space for "Hello".

I do understand that using a fixed length array will work very well.
But I wish to find out how can this be achieved using pointers.

Thank you.

Stan

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

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

        char *Hello = NULL;
        char *World = NULL;
        Hello = malloc(100);
        Hello = "hello";
        World = "world";
        strcat(Hello, World);
        printf("%s\n",Hello);

        return 0; 
}

[Salem]

> Hello = malloc(100);
Ok, you allocated space

> Hello = "hello";
Oops, you just lost it again (and you've just leaked some memory)
You probably wanted strcpy(Hello,"hello");

[Stanley S]

Oh. Silly me. I must copy <somethings> into the pointed location.

Thank you Salem.

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

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

        char *Hello = NULL;
        char *World = NULL;
        Hello = malloc(100);
        strcpy(Hello, "hello");
        World = "world";
        strcat(Hello, World);
        printf("%s\n",Hello);

        return 0; 
}

[Stanley S]

In this case,

Do I have to malloc() every single char pointers? For eg:

World = "world";

If I do not malloc() "World", am I right to say that "World" is just writing somewhere in the memory? (Having a possibility of overwriting some other app's memory space?)

[Salem]

No, the compiler allocates space for "strings" automatically.

So when you do
World = "world";
You're just assigning a pointer to the start of that memory the compiler set up for you.

The reason for the segfault is these "strings" really are constant, and the OS will stop your program if you try and modify them (as your first strcat tried to do).

[dwks]

printf("%s\n",Hello);

->

puts(Hello);

And don't forget to use free().