Array of pointers with malloc

Question

I have to do a task for university c course and I'm stuck (i won't ask you to write it for me, it's just a simple question). I need to:

malloc an array of pointers in which i'll save adresses of memory assinged to strings, which i'll enter
activate function n times (n-number of strings) in which i enter the string and save it to its assigned memory
activate function which sorts (with bubblesort) the array of pointers in alphabetical order of entered strings.
print sorted strings.

i'm stuck here:

void enter(char *pointer[], int b, int l)
{
    char temp[l];
    printf("Enter string: ");
    scanf("%s",temp);
    printf("Test: %s; %d \n",temp, &temp);
    pointer[b]=temp;
    printf("Test: %s; %d \n",pointer[b], &pointer[b]);
}

void druga()
{
    int i, ile=1, leng=100;
    char *wsk[ile];
    for(i=0;i<ile;i++) wsk[i]=(char*)malloc(leng*sizeof(char));
    
 
    for(i=0;i<ile;i++) 
    {
        enter(wsk,  i, leng);
    }
    for(i=0;i<ile;i++) 
    {
        printf("Test2 %s; %d \n", wsk[i], &wsk[i]);
            
    }
    //bubblesort(wsk);
   
}

int main(void)
{
    druga();
    return 0;
}

Output:

Enter string: asd
Test: asd; -402654896 
Test: asd; -402654688
Test2 ; -402654688

The issue is that somehow the string does not pass to druga() function. Does anyone have a solution? Thanks for help

`pointer[b]=temp;` copies the pointer, not the contents of `temp` and `temp` is invalid once the function ends. Research `strdup()` and `strcpy()`. — chux - Reinstate Monica, Jan 04 '22 at 19:46
I attempted running your code myself without changing anything, and my last line of output was `Test2 asd; 1784384112`, seems that the string is being passed fine, maybe try running it in a different environment? — Ewan Brown, Jan 04 '22 at 19:48
[How to printf a memory address in C](https://stackoverflow.com/questions/30354097/how-to-printf-a-memory-address-in-c/30354164) — yano, Jan 04 '22 at 19:55
Your homework glosses over step 0, in which you allocate the space for the strings themselves. — Neil, Jan 04 '22 at 20:11

0___________ · Accepted Answer · 2022-01-04T20:27:05.747

    char temp[l];

is a local automatic storage variable and it cannot be accessed via the pointer after the function exit, as this variable stops existing.

You have many less important issues in your code as well.

void enter(char *pointer[], int b, int l)
{
    char temp[l];
    printf("Enter string: ");
    scanf("%s",temp);
    printf("Test: %s; %p \n",temp, (void *)temp);
    strcpy(pointer[b], temp);
    printf("Test: %s; %p \n",pointer[b], (void *)pointer[b]);
}

void druga()
{
    int i, ile=1, leng=100;
    char *wsk[ile];
    for(i=0;i<ile;i++) wsk[i]=malloc(leng*sizeof(**wsk));
    
 
    for(i=0;i<ile;i++) 
    {
        enter(wsk,  i, leng);
    }
    for(i=0;i<ile;i++) 
    {
        printf("Test2 %s; %p \n", wsk[i], (void *)wsk[i]);
            
    }
    //bubblesort(wsk);
   
}

int main(void)
{
    druga();
    return 0;
}

This program should also free dynamically allocated memory, check for the allocation errors, check the result of scanf and limit number of entered characters to do not write outside the temp array bounds.

Your answers were helpful, thank you both. Code works perfectly — Caligula, Jan 04 '22 at 23:42

Array of pointers with malloc

1 Answers1