Thread: How to set up Dynamic Storage/ Temporary Buffer?

So this is my code so far.

Code:

#include <stdio.h>
#include <stdlib.h>
#define LUNCHES 5
#define ARRAY 2


int main(void)
{
    int x;
    struct Food
    {
        char *name;                                                            /* “name” attribute of food */
        int weight, calories;                                                  /* “weight” and “calories” attributes of food */
    }lunch[LUNCHES] = { [0] = {"apple", 4, 100}, [1] = {"salad", 2, 80} };


    for(x = ARRAY; x <= LUNCHES; ++x)
    {


        printf("Please input \"food\", weight, calories: ");
        scanf("%d %d", &lunch[x].weight, &lunch[x].calories);
        printf("The %s weighs %doz. and contains %d calories", lunch[x].name, lunch[x].weight, lunch[x].calories);


    }
    
    return 0;
}

I need to set a temporary buffer for food. Not sure how to do that. Also allocating memory.

I think for allocating:

Code:

Food *buff = malloc(sizeof(name))

Along the lines of it, what do you think?

You should test it and find out.If you do not understand the behavior of your program post back.

Think of how much space does you need to allocate in your memory to store the struct....

Originally Posted by std10093

You should test it and find out.If you do not understand the behavior of your program post back.

Think of how much space does you need to allocate in your memory to store the struct....

Well I tried this.

Code:

#include <stdio.h>
#include <stdlib.h>
#define LUNCHES 5
#define ARRAY 2


int main(void)
{
    int x;
    struct Food
    {
        char *name;                                                            /* “name” attribute of food */
        int weight, calories;                                                  /* “weight” and “calories” attributes of food */
    }lunch[LUNCHES] = { [0] = {"apple", 4, 100}, [1] = {"salad", 2, 80} };


    for(x = ARRAY; x <= LUNCHES; ++x)
    {
        char *buff = malloc(sizeof(lunch[x].name));


        printf("Please input \"food\", weight, calories: ");
        scanf("%s", buff);
        scanf("%d %d", &lunch[x].weight, &lunch[x].calories);
        printf("The %s weighs %doz. and contains %d calories.\n", lunch[x].name, lunch[x].weight, lunch[x].calories);




    }


    return 0;
}

Shows up as The NULL weighs input and contains input calories

Originally Posted by root

you have to allocate memory for a string, you dont' need buff variable
Try

Code:

lunch[x].name=malloc(50*sizeof(char)); 
/*I'm assuming that the food name is shorter than 50 characters, write  "=malloc(50);" is the same because char is always 1, 
but put it always  and you'll never wrong*/

So that does work. But the last thing is to check for null. I have the following code

Code:

if (lunch[x].name == NULL)        {
            printf("There is an error with the string");
            break;
        }

I think I have to do something like

Code:

char *buff;
if ((buff = (char*)malloc(25 * sizeof(char))) == NULL)
{
    printf("There has been an error with the string %s", lunch[x].name)
    break;
    /* how do I exit? */
}

Originally Posted by Radiant

So that does work. But the last thing is to check for null. I have the following code

Right observation , if you want to exit if the allocation fails you can call the exit function writing

Code:

exit(1);

Cast of what malloc returns is ooooold style. To be exact it is not recommended.

When main wants to say that something went wrong , it returns -1 instead of 0 ,as every main ought to return right before terminating...

Also you might use these
EXIT_FAILURE - C++ Reference
EXIT_SUCCESS - C++ Reference