What does this array error mean?

Question

void arrayRound(int id, double baln)
{
    baln[id] = (baln[id]*100) + 0.5;
    int temp = (int) baln[id];
    baln[id] = (double) temp;
    baln[id] = baln[id] / 100;
}

The function body is what is giving me error messages. The function is meant to round an array index to the nearest hundredth. I separately passed both the index variable and the array to the function. Here is the error message:

Fxns.c:70: error: subscripted value is neither array nor pointer
Fxns.c:70: error: subscripted value is neither array nor pointer
Fxns.c:71: error: subscripted value is neither array nor pointer
Fxns.c:72: error: subscripted value is neither array nor pointer
Fxns.c:73: error: subscripted value is neither array nor pointer
Fxns.c:73: error: subscripted value is neither array nor pointer

My first guess was that I needed to include empty brackets after the baln in the parameter field, but that didn't help. Any ideas?

Answer 1

您正在尝试治疗baln类型的double像一个数组（使用索引）。这是行不通的。

Answer 2

Your parameter should be declared as

double *baln

a pointer to double , or as double baln[] , which reads like an array of double s, but as a function parameter also denotes a pointer.

void arrayRound(int id, double *baln)
{
    baln[id] = (baln[id]*100) + 0.5;
    int temp = (int) baln[id];
    baln[id] = (double) temp;
    baln[id] = baln[id] / 100;
}

will compile, but since you don't know what size the memory block baln points to is, you may access unallocated memory in this function, if you are not careful.

Answer 3

You had it right; you do need to include empty brackets after baln in the parameter list, like so:

void arrayRound(int id, double baln[]);

Here's a full demo.

Answer 4

error: subscripted value is neither array nor pointer

baln[id]

Subscripted Value = baln

Operator [] , can only be used either on arrays or pointers. In your case, baln is neither. It is of type double but not double[] or double* .

int a[] = { 1,2,3,4,5 };
a[0] = 10;  // Notice the use of `[]`.This is valid because `a` is an array type.

int b = 10;
int * ptr = &b;
ptr[0] = 99;   // Valid because ptr is a pointer type but cause undefined
               // behaviour for any other index in this example.

*ptr = 99 ; // This is more readable than the earlier case.