I have written program that suppose to print reverse number for given size
Program is not working as it suppose to do.Code:#include<stdio.h> int reverse ( int *p, int a[5]) { int i = 5; for ( i = 5; i > 0; i--) { p = a[i]; printf("%d ", *p); p++; } } int main () { int result; int i = 0; int a[5] = {1, 2, 3, 4, 5 }; result = reverse (&a, a[5]); return 0; }
What's fault in program ?
I suggest that the function look like this:
Here's why:Code:void print_reverse(const int numbers[], int size)
- print_reverse makes it clear that you're printing the numbers in reverse, not reversing them in-place.
- The return type is void because you're not returning anything.
- The first parameter is named numbers because that's what the array contains. I have used array notion to indicate that we expect to deal with an array, even though you must keep in mind that numbers is a pointer.
- The type of what numbers points to is const int because you only want to print, not to modify.
- The second parameter is named size because that's what you want to print in reverse "for given size".
- size is an int because a size is an integer (but if you have learnt about it, it would be better to make it a size_t).
You would then call the function like this:
The idea is that a is converted to a pointer to its first element, so there's no need to take its address (and in fact that would be wrong because the types would not match). Especially if you chose to make size a size_t, you might compute the size instead of hard coding it:Code:print_reverse(a, 5);
In your original code, passing a[5] is wrong because a[5] does not exist for an array a with 5 elements, and besides you want to pass the size.Code:print_reverse(a, sizeof(a) / sizeof(a[0]));
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
I want to reverse them in place
Imagine numbers are stored as following
0x00 = 1
0x01 = 2
0x02 = 3
0x03 = 4
0x04 = 5
when I reverse them in place
0x00 = 5
0x01 = 4
0x02 = 3
0x03 = 2
0x04 = 1
That's what I want to do
How it can be done ? I was trying to achieve that in my code but I couldn't get result
In that case, I suggest writing two functions. First:
This will print the first size number of numbers. Implement and test that this works, then implement:Code:void print(const int numbers[], int size)
This second function will not print at all. Rather, it will only reverse in-place, and then later you call the print function to print the result.Code:void reverse(int numbers[], int size)
Look up a C++ Reference and learn How To Ask Questions The Smart Way
First, write a test function:
Now write your main function to assert what you want to be true:Code:bool arrays_equal(int *a1, size_t len. int * a2) { if (NULL == a1 || NULL == a2) return len == 0; // equal for 0 elts, otherwise not equal for (int i = 0; i < len; ++i) if (a1[i] != a2[i]) return false; return true; }
Code:#if !defined(__STDC_VERSION__) || __STDC_VERSION__ < 201112L #error "Need C11" #endif #define array_length(a) (sizeof (a) / sizeof (*a)) void reverse_array(int *, size_t); int main(void) { reverse_array(NULL, 0); // These had better do something! reverse_array(NULL, 1); // Write a test case. int a1[] = {1, 2, 3, 4, 5}; reverse_array(a1, array_length(a1)); assert(arrays_equal(a1, array_length(a1), (int[]){5, 4, 3, 2, 1})); int a2[] = {3, 7}; reverse_array(a2, array_length(a2)); assert(arrays_equal(a2, array_length(a2). (int[]){7, 3})); int a3[] = {5}; reverse_array(a3, array_length(a3)); assert(arrays_equal(a3, array_length(a3), (int[]){5})); int a4[] = {}; reverse_array(a4, array_length(a4)); assert(arrays_equal(a4, array_length(a4), (int[]){5})); }
