C语言 -> C 结构的无效类型参数
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Invalid type argument of -> C structs
提问by system
I am trying to access items in an array of structs and print the structs fields as follows
我正在尝试访问结构数组中的项目并按如下方式打印结构字段
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
但我收到了很多错误
invalid type argument of '->'
“->”的无效类型参数
What am I doing wrong with this pointer?
我用这个指针做错了什么？
回答by James McNellis
ais of type Album*which means that a[i]is of type Album(it is the ith element in the array of Albumobject pointed to by a).
ais of typeAlbum*这意味着它a[i]的类型Album（它是指向i的Album对象数组中的第 th 个元素a）。
The left operand of ->must be a pointer; the .operator is used if it is not a pointer.
的左操作数->必须是指针；该.运营商使用，如果它不是一个指针。
回答by J?rgen Sigvardsson
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:
您需要使用.运算符。你看，当你*对一个指针应用 a 时，你是在取消引用它。同理[]。*和之间的区别在于[]括号需要指针的偏移量，在取消引用之前，该偏移量被添加到指针中的地址。基本上，这些表达式是相同的：
*ptr == ptr[0]
*(ptr + 1) == ptr[1]
*(ptr + 2) == ptr[2]
To connect to your question: Change a[i]->field2and a[i]->field3to a[i].field2and a[i].field3.
要连接到你的问题：变革a[i]->field2和a[i]->field3以a[i].field2和a[i].field3。
