错误：在 function 中打印数组时，下标值既不是数组也不是指针也不是向量（C）

Question

Hi there I'm trying to print an array but I keep getting this error out (error: subscripted value is neither array nor pointer nor vector) whenever I try and print from a function. As its a time I want only the data within the array not the position of the data as well.
***It also works if its not in a function.

My code is as follows:

int currenttime[] = {00,00,00};
int i;
void outputtime(int currenttime)
{   
    for(i = 0; i<3; i++) 
    {
     printf("%d ", i, currenttime[i]);
    }
}

int main()
{
outputtime(currenttime);
return 0;
}

Answer 1

You need to provide parameters to match the prototype of a called function.
Either you keep the prototype and call accordingly, or you adapt the prototype.

This is giving a pointer to int (to which the use of an array identifier in a function call decays) to a function which you declared to expect an int.

outputtime(currenttime);

From the context I conclude that in this case your call is actually what you want to do, so the prototype needs to be adapted to declare a parameter of pointer to int .

void outputtime(int* currenttime)

But that gives an output of "1 2 3".
This happens because in your call to printf() you provide two values ( i and currenttime[i] ), with only one being used by the format specifier "%d " .

Either you want to print the two values and need to adapt the specifier

printf("%d:%d, ", i, currenttime[i]);

which gets you an output of

0:0, 1:0, 2:0, 

or you remove the first value to only output the content of the array.

printf("%d ", currenttime[i]);

for an output of

0 0 0 

In case you now wonder why your init of the array of

int currenttime[] = {00,00,00};

only gets you single 0s for each part; that is because the 00 are identical to 0 . If you want to change how a 0 is printed, you need to use other features of printf() , after reading the details in the specification.