Selection Sort in C using an Array of Struct, error: "lvalue required..."

Question

Trying to sort an array of Structs. Struct is TextArt defined below

typedef struct  //struct that holds ASCII art
{
    char artistName[80];    //name of artist
    char asciiArt[20][80];  //actual ascii art line by line
    int rating;             //rating of art
}TextArt;

I don't think the struct has anything to do with this though. I get the compiler error

error: lvalue required as left operand of assignment when trying to assign one struct to another(seen below)

temp = asciiArt+pos_min;
asciiArt+pos_min = asciiArt+i;  //error here
asciiArt+i = *temp;  //error also here

Call to function

selectionSort(artPtr, artArrSize);

and full selection sort function. Is there something I'm not understanding about assigning structs in C using =? I thought it was either this or my passing of the TextArt array was somehow wrong. Please enlighten me and thank you.

void selectionSort(TextArt *asciiArt, int size)
{
    //pos_min is short for position of min
    int pos_min;
    TextArt *temp;
    int i=0;
    int j=0;

    for (i=0; i < size-1; i++)
    {
        pos_min = i;//set pos_min to the current index of array

        for (j = i + 1; j < size; j++)
        {
            if ((strncmp((asciiArt+i)->artistName, (asciiArt+j)->artistName)) < 0)
            {
                pos_min = j; //pos_min will keep track of the index that min is in, this is needed when a swap happens
            }
        }

    //if pos_min no longer equals i than a smaller value must have been found, so a swap must occur
            if (pos_min != i)
            {
                printf("copying...\n");
                temp = asciiArt+pos_min;
                asciiArt+pos_min = asciiArt+i;
                asciiArt+i = *temp;
    }
}

this · Accepted Answer · 2014-04-18 20:26:42Z

2

A correct way to swap the two structs would be:

if (pos_min != i)
{
    printf("copying...\n");
    const TextArt temp = asciiArt[pos_min]; //equivalent to: *(asciiArt + pos_min)
    asciiArt[pos_min] = asciiArt[i];
    asciiArt[i] = temp;
}

answered Apr 18, 2014 at 20:26

this

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5 Comments

lloyd Over a year ago

Tried this code and it's just totally ignored, "copying..." does not even display.

this Over a year ago

If this part of the code was never reached then where is the problem? At a glance the sorting seems correct, only for (j = i + 1 should be changed.

lloyd Over a year ago

I'm not sure, but when I use memcpy, "copying..." is printed, with this code no output. Nothing else in the loop besides printf, no output.

this Over a year ago

Your code has errors, enable compilers warnings and try to compile.

lloyd Over a year ago

got it.... there was a stray character from copying/pasting messing the code up. Thanks~

brokenfoot · Accepted Answer · 2014-04-18 19:58:48Z

0

You cannot copy structs by assigning one pointer to the other. This way you are not creating two copies instead you are pointing both the pointers to the same address. You need to use memcpy instead. This is wrong:

temp = asciiArt+pos_min;

instead, use:

void * memcpy ( void * destination, const void * source, size_t num );

answered Apr 18, 2014 at 19:58

brokenfoot

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4 Comments

lloyd Over a year ago

Thank you. God that was a big headache :)

this Over a year ago

Why complicate with memcpy if you can assing a temporary struct. Its better faster and less error prone.

lloyd Over a year ago

@self can you give me an example of how to do this?

Jonathan Leffler Over a year ago

You can assign structs; you just need to copy whole structs: TextArt temp = asciiArt[min]; asciiArt[min] = asciiArt[i]; asciiArt[i] = temp; works fine.

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