实现一个指针function遍历数组中的元素

Question

I want to traverse the array element with pointer function by C. However, there's some limitations about doing the project.

Here is the array:

ary[5] = {1,2,3,4,5}

I have to create three pointer-int-funciton like as follow:

int* begin();           
int* end();             
int* next();            
void printArray();     

So, when I implement the funciton, I doing the way like this:

//define the begin function
int* begin(){
    int *index = &ary[0];       
    return index;
}
//define the next function
int* next(){
    int *count = ary;
    count ++;
    return count;
}
//define the end function
int* end(){
    int *index = ary;
    while (*index < &ary[5])
    {

        index ++;
    }
}

printArray function:

 void printArray(){
    
        for (begin(); next() <= end(); next())
        {
            printf("%d ",*next());
        }
    }

However, the next() function could not work, because using the upper code will enter an Infinite loop. I would like to code like could++. I hope someone can figure the problem with me.

Answer 1

1. Your declaration of int array should be int ary[5] = {1, 2, 3, 4, 5} , you missed the type.
2. Every time you call next() , you get a copy of ary's address to count , count++ only increment the copied pointer, not the ary itself, so you will print infinite 2 on the screen.
3. You cannot increment the ary because it's an array not a pointer, or you will get error.

A simple way to print an int array like this:

int main(){
    int ary[5] = {1, 2, 3, 4, 5};
    int *ptr = ary;
    while(ptr <= &ary[4]){
        printf("%d ", *ptr);
        ptr++;
    }
    return 0;
}

Modified code of your functions:

int ary[5] = {1,2,3,4,5};
int *ptr;

int* begin();           //set ptr to the address of ary, and return it
int* end();             //the memory space address of the last element of the array
void next();            //increment the ptr
void printArray();      //print out the values in the array

int* begin(){
    ptr = &ary[0];
    return ptr;
}
void next(){
    ptr++;
}
int* end(){
    //return &ary[4];
    return &ary[sizeof(ary)/sizeof(ary[0])-1];
}
void printArray(){
    for (begin(); ptr <= end(); next())
    {
        printf("%d ",*ptr);
    }
}

int main(){
    printArray();
    return 0;
}

Answer 2

Here is my code as below:

int* begin();           
int* end();             
int* next();            
void printArray();
int ary[SIZE] = {},     
    data;               
static int *pointer;

The main function as below:

//Using the for-loop to implement the program
for (int i = 0; i < SIZE; i ++)
{
    //input the number to the array
    printf("Enter the number: ");
    scanf("%d", &data);
    if (data < 0)
    {
        break;

    }else{
        ary[i] = data;
    }
}
//define the begin function
int* begin(){
    pointer = &ary[0];
    return pointer;
}
//define the next function
int* next(){
    pointer ++;
    return pointer;
}

//define the end function
int* end(){
    return &ary[sizeof(ary)/sizeof(ary[0])-1];
}

//define the printArray function
void printArray(){

    for (begin(); pointer <= end(); next())
    {
        printf("%d ",*pointer);
    }
}

If I enter the 3 number, for example 1 2 3, the results will not get the end of the number 3, at the same time, it will print out 1 2 3 0 0. I know that I have to remove the last two element(that doesn't input to the ary). So, I feel a little confused. How can I solve it?