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Invalid type argument of -> C structs

开发者 https://www.devze.com 2023-02-10 18:01 出处:网络
I am trying to access items in an array of开发者_如何转开发 structs and print the structs fields as follows

I am trying to access items in an array of开发者_如何转开发 structs and print the structs fields as follows

printList(Album *a, int numOfStructs)
{
    int i;
    int j;

    for(i = 0; i < numOfStructs; i++)
    {
         printf("number%d\n:", i+1);
         printf("%s", a[i]->field2);
         printf("%s", a[i]->field2);
         printf("%d", a[i]->field3);

         for(j = 0; j < a[i]->numOfStrings; j++)
         {
             printf("%s", a[i]->strings[j]);
         }
         printf("\n");
    }
}

but I get loads of errors as such

invalid type argument of '->'

What am I doing wrong with this pointer?


a is of type Album* which means that a[i] is of type Album (it is the ith element in the array of Album object pointed to by a).

The left operand of -> must be a pointer; the . operator is used if it is not a pointer.


You need to use the . operator. You see, when you apply a * to a pointer, you are dereferencing it. The same goes with the []. The difference between * and [] is that the brackets require an offset from the pointer, which is added to the address in the pointer, before it is dereferenced. Basically, these expressions are identical:

*ptr == ptr[0]
*(ptr + 1) == ptr[1]
*(ptr + 2) == ptr[2]

To connect to your question: Change a[i]->field2 and a[i]->field3 to a[i].field2 and a[i].field3.

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