simple pointer question

i.you · 04-12-2007, 09:55 PM

Hello.

Please look at this code:

int main(int argc, char **argv)
{
...
char **temp;

for(i=0; i<argc, i++) {
// *temp[i] = *argv[i];
strcpy(temp[i], argv[i]);
}
...

The segmentation faults occur in the for loop.
The number of argv is more than 300.
What's wrong?

Please help me.

nadroj · 04-12-2007, 10:05 PM

temp isnt initialized. so your trying to copy argv[i] to a place that hasnt been allocated memory.

i.you · 04-12-2007, 11:13 PM

Thanks.

but the seg faults also occus although allocate memory to temp, i.e.,

char **temp = (char **)malloc(SIZE);

nadroj · 04-12-2007, 11:45 PM

heres an example

Code:

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

int main()
{
    int SIZE = 1;
    int MAX_STR_LEN = 20;
    
    char* word = "hello there";
    
    char** temp = (char**) malloc(sizeof(char*) * SIZE);

    temp[0] = (char*) malloc(sizeof(char) * MAX_STR_LEN);
    
    strcpy(temp[0],word);
    printf("'%s'\n",temp[0]);
    
    free(temp[0]);
    free(temp);
    
    getchar();
}

jschiwal · 04-12-2007, 11:51 PM

Quote:

Originally Posted by nadroj

please note: my post count does not mean i am a guru, i consider myself a linux newbie

The world could use a few more newbies like you!

nadroj · 04-12-2007, 11:56 PM

/deleteme

reddazz · 04-13-2007, 02:29 PM

Moved: This thread is more suitable in the Programming forum and has been moved accordingly to help your thread/question get the exposure it deserves.

kaz2100 · 04-14-2007, 10:33 AM

Hya,

You allocated memory to **yourHandle,

What you have to allocate is *yourPointer.

**temp is a pointer to pointer, allocate memory to *temp. not **temp.

Code:

for(......)   *temp[i] = (char*)malloc(yoursize);

Happy Penguins